Exam Questions Chapter.3 The Mole And Stoichiometry - Solution Bank | Chemistry Molecular Nature 8e by Neil D. Jespersen. DOCX document preview.
Chemistry: Molecular Nature of Matter, 8e (Jespersen)
Chapter 3 The Mole and Stoichiometry
1) An atom of copper has a mass about four times greater than that of an atom of oxygen. What choice makes the correct comparison of the relative numbers of copper and oxygen atoms in 1,000 g of each element?
A) The number of copper and oxygen atoms is the name.
B) There are one thousand times as many copper atoms as oxygen atoms.
C) There are one thousand times as many oxygen atoms as copper atoms.
D) There are four times as many oxygen atoms as copper atoms.
E) There are four times as many copper atoms as oxygen atoms.
Diff: 1
Section: 3.1
2) An atom of sulfur is about four times smaller in mass than an atom of iodine. How many grams of neon will contain the same number of atoms as 1,000 g of iodine?
A) 4 g S
B) 250 g S
C) 400 g S
D) 1,000 g S
E) 4,000 g S
Diff: 1
Section: 3.1
3) Which of these quantities does not represent 1.00 mol of the indicated substance?
A) 6.02 × 1023 C atoms
B) 26.0 g Fe
C) 12.01 g C
D) 65.4 g Zn
E) 6.02 × 1023 Fe atoms
Diff: 1
Section: 3.1
4) If 0.274 moles of a substance weighs 62.5 g, what is the molar mass of the substance, in units of g/mol?
A) 2.28 × 102 g/mol
B) 1.71 × 101 g/mol
C) 4.38 × 10-3 g/mol
D) 2.17 × 102 g/mol
E) none of these
Diff: 1
Section: 3.1
5) The atomic mass of C is 12.011 u. How many moles of C are there in a 3.50 g sample of carbon?
A) 0.291 moles
B) 0.374 moles
C) 1.00 moles
D) 3.43 moles
E) 3.50 moles
Diff: 1
Section: 3.1
6) The atomic mass of Mg is 24.305 u. How many moles of Mg are there in a 3.50 g sample of magnesium?
A) 0.0182 moles
B) 0.144 moles
C) 0.218 moles
D) 0.226 moles
E) 1.31 × 1023 moles
Diff: 1
Section: 3.1
7) The atomic mass of aluminum is 26.982 u. How many moles of Al are there in a 4.56 g sample of aluminum?
A) 0.169 moles
B) 0.220 moles
C) 1.33 moles
D) 4.55 moles
E) 5.93 moles
Diff: 1
Section: 3.1
8) The atomic mass of chromium is 51.996 u. How many moles of Cr are there in a 5.44 g sample of chromium?
A) 0.0875 moles
B) 0.0907 moles
C) 0.105 moles
D) 0.220 moles
E) 2.33 moles
Diff: 1
Section: 3.1
9) The atomic mass of cobalt is 58.933 u. How many moles of Co are there in a 7.60 g sample of cobalt?
A) 0.106 moles
B) 0.114 moles
C) 0.123 moles
D) 0.129 moles
E) 7.79 × 1022 moles
Diff: 1
Section: 3.1
10) The atomic mass of helium is 4.0026 u. What is the mass of a helium sample that contains 0.427 moles of He gas?
A) 0.427 g
B) 0.107 g
C) 1.71 g
D) 2.57 g
E) 9.37 g
Diff: 1
Section: 3.1
11) The atomic mass of cobalt is 58.993 u. What is the mass of a cobalt sample that contains 0.763 moles of cobalt?
A) 39.7 g
B) 40.3 g
C) 43.4 g
D) 45.0 g
E) 45.7 g
Diff: 1
Section: 3.1
12) The atomic mass of boron is 10.811 u. What is the mass of a boron sample that contains 0.585 moles of boron?
A) 0.00541 g
B) 1.80 g
C) 3.52 g
D) 6.32 g
E) 18.5 g
Diff: 1
Section: 3.1
13) The atomic mass of phosphorus is 30.974 u. What is the mass of a phosphorus sample that contains 0.585 moles of phosphorus?
A) 17.3 g
B) 18.1 g
C) 22.3 g
D) 26.5 g
E) 34.2 g
Diff: 1
Section: 3.1
14) The molar mass of Co(NH3)6(ClO4)3 is
A) 318.53 u.
B) 389.43 u.
C) 402.57 u.
D) 459.47 u.
E) 754.13 u.
Diff: 2
Section: 3.1
15) The molar mass of C14H28(COOH)2 is
A) 242.40 u.
B) 254.41 u.
C) 270.41 u.
D) 273.39 u.
E) 286.41 u.
Diff: 2
Section: 3.1
16) The molar mass of Ni(H2O)6Cl2 is
A) 157.69 u.
B) 193.00 u.
C) 227.61 u.
D) 237.69 u.
E) 296.83 u.
Diff: 2
Section: 3.1
17) The molar mass of (NH4)3PO4 is
A) 116.03 u.
B) 121.07 u.
C) 149.09 u.
D) 155.42 u.
E) 242.01 u.
Diff: 2
Section: 3.1
18) The molar mass of (NH4)2SO4 is
A) 84.12 u.
B) 116.12 u.
C) 118.13 u.
D) 132.14 u.
E) 221.53 u.
Diff: 1
Section: 3.1
19) A gas sample contains 16.0 g of CH4, 16.0 g of O2, 16.0 g of SO2, and 33.0 g of CO2. What is the total number of moles of gas in the sample?
A) 2.25 moles
B) 2.50 moles
C) 2.75 moles
D) 3.00 moles
E) 4.00 moles
Diff: 2
Section: 3.1
20) The mass of 5.20 moles of glucose, C6H12O6 is
A) 1.56 × 1021 g.
B) 31.2 g.
C) 34.7 g.
D) 937 g.
E) 6.43 × 1020 g.
Diff: 2
Section: 3.1
21) If the atomic mass of gold is 196.9665 u, how many grams of gold are in 0.150 mol Au?
A) 7.62 × 10-4 g
B) 29.5 g
C) 29.54498 g
D) 7.61551 × 10-4 g
E) 0.903 g
Diff: 1
Section: 3.1
22) How many atoms of 12C are in a 3.50 g sample of this particular isotope?
A) 1.52 × 1023
B) 1.75 × 1023
C) 2.07 × 1024
D) 2.11 × 1024
E) 8.01 × 1023
Diff: 2
Section: 3.1
23) The atomic mass of aluminum is 26.982 u. How many aluminum atoms are in a 4.55 g sample of aluminum?
A) 1.02 × 1023
B) 1.32 × 1023
C) 2.74 × 1024
D) 3.57 × 1024
E) 8.01 × 1023
Diff: 2
Section: 3.1
24) The atomic mass of neon is 20.1797 u. What is the mass, in grams, of a neon sample which contains 1.00 × 1020 neon atoms?
A) 1.66 × 10-4 g
B) 2.02 × 10-19 g
C) 298 g
D) 3.35 × 10-3 g
E) 8.33 × 10-6 g
Diff: 2
Section: 3.1
25) The atomic mass of magnesium is 24.3050 u. What is the mass, in grams, of a magnesium sample which contains 3.22 × 1022 magnesium atoms?
A) 1.30 g
B) 12.2 g
C) 1.99 g
D) 5.88 g
E) 7.50 × 10-4 g
Diff: 2
Section: 3.1
26) How many molecules of carbon dioxide are in 154.0 grams of carbon dioxide?
Hint: Think about your pathway to solving. Can you convert directly from mass to molecules or is there an intermediate unit you should use?
A) 3.499
B) 2.107 × 1024
C) 4.214 × 1024
D) 9.274 × 1025
E) 4.081 × 1027
Diff: 3
Section: 3.1
27) How many formula units of sodium chloride are in 146 grams of sodium chloride?
Hint: Think about your pathway to solving. Can you convert directly from mass to formula units or is there an intermediate unit you should use?
A) 3.88 × 10-22
B) 2.29 × 1024
C) 1.50 × 1024
D) 7.27 × 1025
E) 4.08 × 1023
Diff: 3
Section: 3.1
28) How many moles of carbon atoms are in 0.145 moles of the compound, C3H8?
A) 2.90 moles
B) 0.435 moles
C) 0.145 moles
D) 6.02 × 1023 moles
E) 29.9
Diff: 1
Section: 3.2
29) How many moles of oxygen atoms are in 1.08 moles of Ca(NO3)2?
A) 7.55 moles
B) 1.43 moles
C) 6.48 moles
D) 33.8 moles
E) 1.16 × 1023 moles
Diff: 1
Section: 3.2
30) A sample of Ca3(PO4)2 contains 3.51 moles of calcium ions. How many moles of Ca3(PO4)2 are in that sample?
A) 3.55 moles
B) 0.491 moles
C) 10.5 moles
D) 1.17 moles
E) 3.51 × 1021 moles
Diff: 1
Section: 3.2
31) How many moles of carbon atoms are combined with 11.2 moles of hydrogen atoms in a sample of the compound, C3H8?
A) 3.00
B) 5.60
C) 4.20
D) 6.02 × 1023
E) 29.9
Diff: 1
Section: 3.2
32) A sample of C12H22O11 contains 0.4662 moles of carbon atoms. How many moles of hydrogen atoms are in the sample?
A) 0.2543 moles
B) 0.4662 moles
C) 10.26 moles
D) 0.8547 moles
E) 0.9324 moles
Diff: 1
Section: 3.2
33) A sample of HOOCC6H4OH contains 0.6540 moles of hydrogen atoms. How many moles of carbon atoms are in the sample?
A) 0.6543 moles
B) 6.090 moles
C) 7.015 moles
D) 1.290 moles
E) 0.7630 moles
Diff: 2
Section: 3.2
34) A sample of Ni(CO)4, a toxic transition-metal complex, has 5.23 × 1024 atoms of carbon. How many atoms of Ni does it contain?
A) 6.02 × 1023 atoms
B) 1.50 × 1023 atoms
C) 4.67 × 1022 atoms
D) 20.9 × 1023 atoms
E) 1.31 × 1024 atoms
Diff: 2
Section: 3.2
35) How many sodium atoms are in 6.0 g of Na3N?
A) 3.6 × 1024 atoms
B) 4.6 × 1022 atoms
C) 1.3 × 1023 atoms
D) 0.217 atoms
E) 0.072 atoms
Diff: 2
Section: 3.2
36) A sample of K3Fe(CN)6 contains 1.084 × 1024 carbon atoms. How many potassium atoms are in the sample?
A) 1.084 × 1024 atoms
B) 3.252 × 1024 atoms
C) 2.168 × 1024 atoms
D) 3.613 × 1023 atoms
E) 5.420 × 1023 atoms
Diff: 2
Section: 3.2
37) A sample of phosphorus trifluoride, PF3, contains 1.400 moles of the substance. How many atoms are in the sample?
A) 4
B) 5.6
C) 8.431 × 1023
D) 2.409 × 1024
E) 3.372 × 1024
Diff: 2
Section: 3.2
38) A sample of arabinose, C5H10O5, contains 0.6000 moles of the substance. How many carbon atoms are in the sample?
A) 3
B) 5
C) 3.613 × 1023
D) 1.807 × 1024
E) 3.011 × 1024
Diff: 2
Section: 3.2
39) Which contains the greatest number of carbon atoms?
A) 0.250 moles of glucose, C6H12O6
B) 1.20 moles of carbon dioxide, CO2
C) 0.500 moles of CaC2
D) 0.450 moles of Al2(CO3)3
E) 0.350 moles of C4H8O2S
Diff: 2
Section: 3.2
40) Which contains the greatest number of hydrogen atoms?
A) 0.300 moles of ammonia, NH3
B) 0.080 moles of sucrose, C12H22O11
C) 0.500 moles of ethanol, C2H5OH
D) 0.250 moles of glucose, C6H12O6
E) 1.30 moles of water, H2O
Diff: 2
Section: 3.2
41) A sample of (N2H5)2C3H4O4 contains 1.084 × 1024 carbon atoms. How many moles of hydrogen atoms are in the sample?
Hint: You can't convert directly between atoms of carbon and moles of hydrogen, use an intermediary unit.
A) 4.200 moles
B) 4.725 moles
C) 7.000 moles
D) 8.400 moles
E) 2.400 moles
Diff: 3
Section: 3.2
42) A sample of sulfolane, C4H8O2S, contains 5.00 × 1024 atoms. How many moles of sulfolane are in the sample?
A) 0.120 moles
B) 0.554 moles
C) 1.81 moles
D) 8.30 moles
E) 3.33 × 1023 moles
Diff: 2
Section: 3.2
43) A sample of trifluoromethanesulfonic acid, CHF3O3S, contains 4.62 × 1023 oxygen atoms. How many moles of CHF3O3S are in the sample?
A) 0.256 moles
B) 2.30 moles
C) 0.259 moles
D) 0.767 moles
E) 1.53 × 1023 moles
Hint: You can't convert directly between atoms of oxygen and moles of CHF3O3S, use an intermediary unit.
Diff: 3
Section: 3.2
44) A sample of C7H5N3O4 contains 0.4662 moles of carbon atoms. How many nitrogen atoms are in the sample?
Hint: You can't convert directly between moles of carbon atoms and atoms of nitrogen, use an intermediary unit.
A) 2.807 × 1023 atoms
B) 1.203 × 1023 atoms
C) 4.101 × 1023 atoms
D) 1.998 × 1022 atoms
E) 6.551 × 1023 atoms
Diff: 3
Section: 3.2
45) How many hydrogen atoms are in 100.00 grams of water (H2O)?
Hint: Use moles as an intermediary unit and keep in mind you need to find hydrogen atoms.
A) 2
B) 200
C) 11.10
D) 6.681 × 1024
E) 3.340 × 1024
Diff: 3
Section: 3.2
46) How many oxygen atoms are in 3.62 g of fructose, C6H12O6?
Hint: Use moles as an intermediary unit and keep in mind you need to find oxygen atoms.
A) 1.56 × 1021
B) 7.26 × 1022
C) 3.45 × 1024
D) 1.95 × 1023
E) 6.44 × 1020
Diff: 3
Section: 3.2
47) How many grams of oxygen are needed to combine with 0.85 g of carbon in chlorophyll, C55H72MgN4O5?
Hint: Use moles as an intermediary unit and focus on the ratio of carbon to oxygen in the formula.
A) 1.56 × 1021 g
B) 0.10 g
C) 0.85 g
D) 1.5 g
E) 6.4 g
Diff: 3
Section: 3.2
48) How many grams of carbon are required to combine with 1.99 g of oxygen in C2H5OH?
Hint: Use moles as an intermediary unit and focus on the ratio of carbon to oxygen in the formula.
A) 5.99 g
B) 4.99 g
C) 1.99 g
D) 2.99 g
E) 3.99 g
Diff: 3
Section: 3.2
49) A sample of C7H5N3O4 has a mass of 7.81 g. What is the mass of oxygen in this sample?
Hint: Use moles as an intermediary unit and focus on the ratio of how many oxygens there are per formula unit.
A) 31.2 g
B) 2.56 g
C) 3.20 × 1023 g
D) 64.0 g
E) 1.75 g
Diff: 3
Section: 3.2
50) What is the percent, by mass, of calcium in Ca(OCl)2?
A) 28.030%
B) 28.571%
C) 31.562%
D) 43.787%
E) 44.493%
Diff: 1
Section: 3.3
51) What is the percent, by mass, of chromium in K2CrO4?
A) 26.776 %
B) 31.763 %
C) 40.268 %
D) 42.241 %
E) 51.996 %
Diff: 1
Section: 3.3
52) What is the percent, by mass, of tungsten in W(CH3)6?
A) 4.3819 %
B) 26.292 %
C) 67.083 %
D) 28.502 %
E) 35.722 %
Diff: 1
Section: 3.3
53) What is the percent, by mass, of boron in Al(BF4)3?
A) 9.501 %
B) 9.385 %
C) 10.152 %
D) 11.285 %
E) 23.713 %
Diff: 2
Section: 3.3
54) What is the percent, by mass, of oxygen in NiSO4∙7H2O?
A) 14.846 %
B) 39.875 %
C) 43.273 %
D) 49.531 %
E) 62.661 %
Diff: 2
Section: 3.3
55) What is the percent, by mass, of nitrogen in NH3?
A) 93.29 %
B) 5.92 %
C) 82.25 %
D) 25.00 %
E) 7.20 %
Diff: 2
Section: 3.3
56) How many grams of nitrogen are present in 100.0 g of NH3?
A) 93.29 g
B) 5.92 g
C) 82.25 g
D) 25.00 g
E) 7.20 g
Diff: 2
Section: 3.3
57) Which one of the following is definitely not an empirical formula?
A) C12H16O3
B) C12H22O11
C) C3H8O2
D) C4H12N2O
E) C6H12O4
Diff: 1
Section: 3.4
58) Which one of the following is definitely not an empirical formula?
A) NH3
B) N2H4
C) Al(NO3)3
D) H3PO4
E) H2O
Diff: 1
Section: 3.4
59) A compound was discovered with a composition by mass of 85.6% C and 14.4% H. Which of these choices could be the molecular formula of this compound?
A) CH4
B) C2H4
C) C3H4
D) C2H6
E) C3H8
Diff: 2
Section: 3.4
60) A compound has an empirical formula of CH2O. An independent analysis gave a value of 150.13 g for its molar mass. What is the molecular formula of the compound?
A) C5H10O5
B) C6H12O6
C) C11H2O
D) C6H6O8
E) C9H10O2
Diff: 1
Section: 3.4
61) A compound has an empirical formula of C2H4O. An independent analysis gave a value of 
for its molar mass. What is the molecular formula of the compound?
A) C4H4O5
B) C10H12
C) C7O3
D) C6H12O3
E) C4H8O5
Diff: 1
Section: 3.4
62) A compound has an empirical formula of CH2. An independent analysis gave a value of 70 g for its molar mass. What is the molecular formula of the compound?
A) C2H4
B) C3H6
C) C4O8
D) C5H10
E) C5H11
Diff: 1
Section: 3.4
63) A compound has an empirical formula of CH2Cl. An independent analysis gave a value of 99.0 g for its molar mass. What is the molecular formula of the compound?
A) CH2Cl
B) C2H4Cl2
C) C2H2Cl4
D) C3H6Cl3
E) C3H3Cl6
Diff: 1
Section: 3.4
64) A compound has an empirical formula of CHCl. An independent analysis gave a value of 194 g for its molar mass. What is the molecular formula of the compound?
A) CHCl
B) C2H2Cl2
C) C3H3Cl3
D) C4H4Cl4
E) C4H3Cl4
Diff: 1
Section: 3.4
65) A 7.300 gram sample of aluminum was quantitatively combined with a sample of selenium to form a compound. The compound had a mass of 39.35 grams. What is the empirical formula for the compound?
A) AlSe
B) Al2Se
C) AlSe2
D) Al2Se3
E) Al3Se2
Diff: 2
Section: 3.4
66) Magnetite is a binary compound containing only iron and oxygen. The percent, by weight, of iron is 72.360%. What is the empirical formula of magnetite?
A) FeO
B) FeO2
C) Fe3O4
D) Fe2O3
E) Fe2O5
Diff: 2
Section: 3.4
67) A 6.789 g sample of a hydrocarbon, upon combustion analysis, yielded 9.883 grams of carbon dioxide. What is the percent, by mass, of carbon in the hydrocarbon?
A) 18.75 %
B) 39.73 %
C) 68.69 %
D) 45.57 %
E) 34.44 %
Diff: 2
Section: 3.4
68) The empirical formula for a compound was determined to be C2H3O2. Which of the following could be the molecular formula for the compound?
A) C2H6O2
B) C4H6O2
C) C3H4O3
D) C6H9O6
E) C8H10O8
Diff: 1
Section: 3.4
69) A 6.987 g sample of a hydrocarbon, upon combustion analysis, yielded 8.398 grams of carbon dioxide. What is the percent, by mass, of carbon in the hydrocarbon?
A) 16.01 %
B) 23.66 %
C) 32.80 %
D) 34.91 %
E) 58.68 %
Diff: 2
Section: 3.4
70) A 4.626 g sample of a hydrocarbon, upon combustion analysis, yielded 6.527 grams of water. What is the percent, by mass, of hydrogen in the hydrocarbon?
A) 14.11 %
B) 15.79 %
C) 41.09 %
D) 66.22 %
E) 85.89 %
Diff: 2
Section: 3.4
71) A 4.266 g sample of a hydrocarbon, upon combustion analysis, yielded 5.672 grams of water. What is the percent, by mass, of hydrogen in the hydrocarbon?
A) 7.44 %
B) 8.62 %
C) 14.88 %
D) 17.24 %
E) 20.07 %
Diff: 2
Section: 3.4
72) A 6.789 g sample of a hydrocarbon, upon combustion analysis, yielded 6.527 grams of water. What is the percent, by mass, of hydrogen in the hydrocarbon?
A) 9.615 %
B) 10.76 %
C) 28.00 %
D) 38.72 %
E) 58.53 %
Diff: 2
Section: 3.4
73) A 6.789 g sample of a compound was analyzed for nitrogen. In the procedure, all of the nitrogen present was completely converted to ammonia (NH3). 1.637 grams of ammonia were obtained. What is the percent, by mass, of nitrogen in the compound?
A) 35.57 %
B) 24.11 %
C) 75.89 %
D) 31.77 %
E) 19.83 %
Diff: 2
Section: 3.4
74) A 7.689 g sample of a compound was analyzed for nitrogen. In the procedure, all of the nitrogen present was completely converted to ammonia (NH3). 1.763 grams of ammonia were obtained. What is the percent, by mass, of nitrogen in the compound?
A) 4.07 %
B) 14.97 %
C) 17.51 %
D) 18.86 %
E) 22.93 %
Diff: 2
Section: 3.4
75) A 4.927 g sample of a compound was analyzed for nitrogen. In the procedure, all of the nitrogen present was completely converted to ammonia (NH3). 1.369 grams of ammonia were obtained. What is the percent, by mass, of nitrogen in the compound?
A) 27.79 %
B) 38.48 %
C) 22.85 %
D) 61.52 %
E) 36.71 %
Diff: 2
Section: 3.4
76) A 7.294 g sample of a compound was analyzed for nitrogen. In the procedure, all of the nitrogen present was completely converted to ammonia (NH3). 3.196 grams of ammonia were obtained. What is the percent, by mass, of nitrogen in the compound?
A) 7.78 %
B) 33.17 %
C) 36.04 %
D) 43.82 %
E) 44.16 %
Diff: 2
Section: 3.4
77) A 3.06 gram sample of an unknown hydrocarbon with empirical formula CH2O was found to
contain 0.0170 moles of the substance. What are the molecular mass and molecular formula, respectively, of the compound?
A) 1.80 × 102 g/mol, C6H12O6
B) 30.1 g/mol, CH2O
C) 5.42 × 103 g/mol, C30H60O30
D) 3.06 g/mol, CH2O
E) 1.80 × 102 g/mol, CH2O
Diff: 2
Section: 3.4
78) A compound contains arsenic and oxygen as its only elements. A 1.626 g sample of the compound contains 1.060 g of arsenic. What is the empirical formula of the compound?
A) As2O
B) AsO
C) As2O3
D) AsO2
E) As2O5
Diff: 2
Section: 3.4
79) A compound contains arsenic and oxygen as its only elements. A 1.626 g sample of the compound contains 1.232 g of arsenic. What is the empirical formula of the compound?
A) As2O
B) AsO
C) As2O3
D) AsO2
E) As2O5
Diff: 2
Section: 3.4
80) A compound contains carbon, hydrogen, and nitrogen as its only elements. A 9.353 g sample of the compound contains 5.217 g of carbon and 1.095 g of hydrogen; the remainder of the mass is nitrogen. What is the empirical formula of the compound?
Hint: Use moles to relate the mass of one element to another.
A) C2H5N
B) C3H5N
C) C3H6N
D) C3H7N
E) C4H9N2
Diff: 3
Section: 3.4
81) A compound contains sodium, boron, and oxygen. An experimental analysis gave values of 53.976 % sodium and 8.461 % boron, by weight; the remainder of the mass is oxygen. What is the empirical formula of the compound?
Hint: Assume you have 100 grams of this substance and use moles to relate the mass of one element to another.
A) NaBO2
B) Na3BO3
C) Na3BO2
D) NaB3O
E) Na3B3O8
Diff: 3
Section: 3.4
82) A compound contains potassium, nitrogen, and oxygen. An experimental analysis gave values of 45.942 % potassium and 16.458 % nitrogen, by weight; the remainder is oxygen. What is the empirical formula of the compound?
Hint: Assume you have 100 grams of this substance and use moles to relate the mass of one element to another.
A) KNO2
B) KNO3
C) K2N2O5
D) KN3O8
E) K2N2O
Diff: 3
Section: 3.4
83) A compound used in a polymer research project contains carbon, hydrogen, and nitrogen. The composition of the compound is: carbon, 58.774%; hydrogen, 13.810%; nitrogen, 27.416%. Determine the empirical formula of the compound.
Hint: Assume you have 100 grams of this substance and use moles to relate the mass of one element to another.
A) C3H7N
B) C2H7N
C) C2H8N
D) C5H14N2
E) C7H16N4
Diff: 3
Section: 3.4
84) A compound contains potassium, sulfur, and oxygen. The composition of the compound is: potassium, 49.410%; sulfur, 20.261%. Determine the empirical formula of the compound.
Hint: Assume you have 100 grams of this substance and use moles to relate the mass of one element to another.
A) K2SO3
B) K2SO4
C) K2S2O4
D) K2S2O3
E) K3S2O8
Diff: 3
Section: 3.4
85) A new compound contains nitrogen, hydrogen, boron, and fluorine. The composition of the compound is: nitrogen, 13.360%; hydrogen, 3.8455%; boron, 10.312%. Determine the empirical formula of the compound.
Hint: Assume you have 100 grams of this substance and use moles to relate the mass of one element to another.
A) NH3BF3
B) NH4B3F
C) N4HB4F
D) NH4BF4
E) NH3BF4
Diff: 3
Section: 3.4
86) A freshly prepared compound contains potassium, hydrogen, phosphorus, and oxygen. The composition of the compound is: potassium, 44.895%; hydrogen, 0.5787%; phosphorus, 17.783%. Determine the empirical formula of the compound.
Hint: Assume you have 100 grams of this substance and use moles to relate the mass of one element to another.
A) K2H4P4O
B) K2HPO4
C) K2H2PO4
D) K2HP2O5
E) KH2PO4
Diff: 3
Section: 3.4
87) A 4.626 g sample of a hydrocarbon, upon combustion analysis, yielded 6.484 grams of carbon dioxide. What is the percent, by mass, of carbon in the hydrocarbon?
Hint: Use moles to relate the quantities of the two substances.
A) 38.25 %
B) 19.47 %
C) 71.35 %
D) 40.16 %
E) 42.16 %
Diff: 3
Section: 3.4
88) In a quantitative analysis study, 4.624 grams of a hydrocarbon compound yielded 13.84 g of CO2 and 7.556 g of H2O in a combustion analysis apparatus. Determine the empirical formula of the compound.
Hint: Use moles to relate the quantities of the substances.
A) CH3
B) C2H5
C) C3H8
D) C10H26
E) C10H27
Diff: 3
Section: 3.4
89) In a quantitative analysis study, 2.644 grams of a hydrocarbon compound yielded 8.008 g of CO2 and 4.098 g of H2O in a combustion analysis apparatus. Determine the empirical formula of the compound.
Hint: Use moles to relate the quantities of the substances.
A) CH3
B) CH4
C) C2H3
D) C2H5
E) C3H8
Diff: 3
Section: 3.4
90) In a quantitative analysis study, 4.624 grams of a compound containing carbon, hydrogen and oxygen yielded 6.557 g of CO2 and 4.026 g of H2O in a combustion analysis apparatus. Determine the empirical formula of the compound.
Hint: Use moles to relate the quantities of the substances.
A) CH2O
B) CH3O
C) CH4O
D) C2H4O
E) C4H2O
Diff: 3
Section: 3.4
91) In a quantitative analysis study, 1.878 grams of a compound containing carbon, hydrogen and oxygen yielded 3.258 g of CO2 and 1.778 g of H2O in a combustion analysis apparatus. Determine the empirical formula of the compound.
Hint: Use moles to relate the quantities of the substances.
A) CH2O
B) C2H3O2
C) C3H5O2
D) C3H7O2
E) C3H8O2
Diff: 3
Section: 3.4
92) Magnesium reacts with aqueous hydrochloric acid to give an aqueous solution of magnesium chloride and hydrogen gas. Select the correct balanced chemical equation for this reaction.
A) Mg(s) + HCl(aq) → 3MgCl(aq) + H2(g)
B) Mg(s) + HCl(g) → MgCl(aq) + H(g)
C) Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
D) 2Mg(s) + 4HCl(aq) → 2MgCl2(aq) + H2(g)
E) 2Mg(s) + HCl(aq) → 2MgCl2(aq) + H2(g)
Diff: 1
Section: 3.5
93) Select the correct balanced chemical equation for the reaction below.
C6H14 + O2 → CO2 + H2O
A) 2C6H14 + 9O2 → 12CO2 + 7H2O
B) 2C6H14 + 19O2 → 12CO2 + 14H2O
C) 2C6H14 + 12O2 → 12CO2 + 14H2O
D) 3C6H14 + O2 → 18CO2 + 22H2O
E) C6H14 + O2 → CO2 + H2O
Diff: 1
Section: 3.5
94) When BaCl2 reacts with Na3PO4, the products formed are Ba3(PO4)2 and NaCl. How many moles of Ba3PO4 are formed for each mole of BaCl2 that is consumed?
A) 3
B) 1
C) 0.3333
D) 2.3333
E) 1.5
Diff: 2
Section: 3.5
95) Given: 3H2(g) + N2(g) s 2NH3(g)
If the reaction begins with 0.500 mol of H2, how many atoms of hydrogen in the compound NH3 would you expect to make?
A) 3.01 × 1023 atoms
B) 6.02 × 1023 atoms
C) 12.04 × 1023 atoms
D) 1 atom
E) 6 atoms
Diff: 2
Section: 3.5
96) Select the correct balanced chemical equation.
Hint: Distribute coefficients across the entire formula they apply to and be sure to correctly calculate the number of atoms inside the parentheses.
A) C2H5OH + 2Na2Cr2O7 + 8H2SO4 → HC2H3O2 + 2Cr2(SO4)3 + 2Na2SO4 + 11H2O
B) 3C2H5OH + 2Na2Cr2O7 + 8H2SO4 → 3HC2H3O2 + 2Cr2(SO4)3 + 2Na2SO4 + 11H2O
C) 2C2H5OH + 2Na2Cr2O7 + 8H2SO4 → 2HC2H3O2 + 2Cr2(SO4)3 + 4Na2SO4 + 11H2O
D) 2C2H5OH + Na2Cr2O7 + 8H2SO4 → 3HC2H3O2 + 2Cr2(SO4)3 + 2Na2SO4 + 11H2O
E) C2H5OH + Na2Cr2O7 + 2H2SO4 → HC2H3O2 + Cr2(SO4)3 + 2Na2SO4 + 11H2O
Diff: 3
Section: 3.5
97) In the chemical reaction C2H6O + PCl3 → C2H5Cl + H3PO3, when the equation is balanced, the sum of the coefficients of the reactants and products should be:
Hint: Distribute coefficients across the entire formula they apply to when calculating atoms.
A) 4
B) 5
C) 6
D) 7
E) 8
Diff: 3
Section: 3.5
98) In the chemical reaction C2H6O + SOCl2 → C2H5Cl + H2SO3, when the equation is balanced, the sum of the coefficients of the reactants and products should be:
Hint: Distribute coefficients across the entire formula they apply to when calculating atoms.
A) 4
B) 5
C) 6
D) 7
E) 8
Diff: 3
Section: 3.5
99) Given the following chemical equation: S + NaOH → Na2SO3 + Na2S + H2O,
the coefficient in front of NaOH should be:
Hint: Distribute coefficients across the entire formula they apply to when calculating atoms.
A) 6
B) 2
C) 5
D) 3
E) 4
Diff: 3
Section: 3.5
100) What is the coefficient of H2O when the following equation is properly balanced with the smallest set of whole numbers?
____ Na + ____ H2O → ____ NaOH + ____ H2
A) 1
B) 2
C) 3
D) 4
E) 5
Diff: 2
Section: 3.5
101) When ethene (C2H4) is burned in the presence of oxygen gas, the products are carbon dioxide and water. When balanced with the smallest set of whole numbers, the coefficient for the oxygen gas molecule is
A) 1.
B) 2.
C) 3.
D) 4.
E) 6.
Diff: 2
Section: 3.5
102) Thermal decomposition of KClO3(s) yields KCl(s) and O2(g). When 4.289 grams of KClO3 undergo this reaction, how many grams of oxygen gas are produced?
Hint: Write and balance the equation first, then do the stoichiometric calculations.
A) 1.120 grams
B) 0.5601 grams
C) 2.240 grams
D) 1.680 grams
E) 4.288 grams
Diff: 3
Section: 3.5
103) When aluminum metal reacts with HCl(aq), the products formed are AlCl3(aq) and the hydrogen gas molecule. If 4.288 grams of Al (0.1590 moles) undergo this reaction with an excess of hydrochloric acid, how many grams of hydrogen gas should be produced?
Hint: Write and balance the equation first, then do the stoichiometric calculations.
A) 0.1603 grams
B) 0.4770 grams
C) 6.048 grams
D) 1.388 grams
E) 0.4808 grams
Diff: 3
Section: 3.5
104) You are given the balanced chemical equation:
C3H8 + 5 O2 → 3 CO2 + 4 H2O
If 0.3818 moles of C3H8 and 1.718 moles of O2 are allowed to react, and this is the only reaction which occurs, theoretically how many moles of water should be produced?
A) 1.336 moles
B) 1.374 moles
C) 1.527 moles
D) 1.718 moles
E) 3.426 moles
Diff: 2
Section: 3.6
105) You are given the balanced equation:
C4H4 + 5 O2 → 4 CO2 + 2 H2O
If 0.3618 moles of C4H4 are allowed to react with 1.818 moles of O2, and this is the only reaction which occurs, what is the maximum quantity of carbon dioxide that could be produced?
A) 1.447 moles
B) 1.454 moles
C) 1.456 moles
D) 2.180 moles
E) 0.3978 moles
Diff: 2
Section: 3.6
106) You are given the balanced chemical equation:
4AsF3 + 3C2Cl6 → 4AsCl3 + 3C2Cl2F4
If 1.3618 moles of AsF3 are allowed to react with 1.000 mole of C2Cl6, what would be the theoretical yield of AsCl3, in moles?
A) 0.3618 moles
B) 0.7343 moles
C) 0.7500 moles
D) 1.3333 moles
E) 1.3618 moles
Diff: 2
Section: 3.6
107) You are given the balanced chemical equation:
C4H4 + 5 O2 → 4 CO2 + 2 H2O
If 0.3618 moles of C4H4 are allowed to react with 1.818 moles of O2, and this is the only reaction which occurs, what is the maximum mass of water that could be produced?
A) 11.02 g
B) 13.04 g
C) 13.20 g
D) 19.64 g
E) 65.50 g
Diff: 2
Section: 3.6
108) Calcium hydroxide and sulfuric acid can react to produce calcium bisulfate and water. When 0.0720 moles of sulfuric acid are reacted with 0.0240 moles of calcium hydroxide, how much calcium bisulfate, Ca(HSO4)2, will be produced?
A) 0.0120 moles
B) 0.0240 moles
C) 0.0360 moles
D) 0.0480 moles
E) 0.0720 moles
Diff: 2
Section: 3.6
109) 150.0 grams of AsF3 were reacted with 180.0 g of CCl4 to produce AsCl3 and CCl2F2. The theoretical yield of CCl2F2 produced, in moles, is
A) 0.7802 moles.
B) 0.5685 moles.
C) 1.274 moles.
D) 1.170 moles.
E) 1.705 moles.
Diff: 2
Section: 3.6
110) The left side of a balanced chemical equation is shown below:
K2Cr2O7 + 4 H2SO4 + 3 SeO2 →
If 0.600 moles of K2Cr2O7, 2.800 moles of H2SO4 and 1.500 moles of SeO2 are brought together and allowed to react, then
Hint: Don't forget to include the coefficients in your consideration.
A) H2SO4 is the limiting reagent.
B) K2Cr2O7 is the limiting reagent.
C) there are 1.300 moles of H2SO4 in excess.
D) there are 0.100 moles of K2Cr2O7 in excess.
E) there are 0.300 moles of SeO2 in excess.
Diff: 3
Section: 3.6
111) Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor:
4NH3 + 5O2 → 4NO + 6H2O
When 40.0 g NH3 and 50.0 g O2 are allowed to react, which is the limiting reagent?
A) NH3
B) O2
C) NO
D) H2O
E) There is no limiting reagent.
Diff: 1
Section: 3.6
112) Chlorine gas can be made from the reaction of manganese dioxide with hydrochloric acid. Which is the limiting reagent when 28 g of MnO2 are mixed with 42 g of HCl?
MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l) + Cl2(g)
Hint: Don't forget to include the coefficients in your consideration.
A) MnO2
B) HCl
C) MnCl2
D) H2O
E) Cl2
Diff: 3
Section: 3.6
113) You are given the unbalanced chemical equation:
C4H8 + O2 → CO2 + H2O
If 0.3218 moles of C4H8 are allowed to react with 2.000 moles of O2, what would be the theoretical yield of water, in moles?
Hint: Don't forget to include the coefficients from the balanced reaction in your consideration.
A) 1.333 moles
B) 1.609 moles
C) 0.6436 moles
D) 1.287 moles
E) 2.574 moles
Diff: 3
Section: 3.6
114) 150.0 grams of AsF3 were reacted with 180.0 g of CCl4 to produce AsCl3 and CCl2F2. The theoretical yield of CCl2F2, in grams, is
Hint: Don't forget to include the coefficients from the balanced reaction in your consideration.
A) 141.4 grams.
B) 206.2 grams.
C) 137.5 grams.
D) 104.4 grams.
E) 152.6 grams.
Diff: 3
Section: 3.6
115) 85.0 grams of H2S were reacted with 115.0 g of O2 to produce SO2 and H2O. The theoretical yield of SO2, in grams, is
Hint: Don't forget to include the coefficients from the balanced reaction in your consideration.
A) 70.5 grams.
B) 230.0 grams.
C) 153.5 grams.
D) 85.0 grams.
E) 159.5 grams.
Diff: 3
Section: 3.6
116) PI3 (411.69 g mol-1) and water (18.015 g mol-1) react to form H3PO3 (81.996 g mol-1) and HI (127.91 g mol-1). If 0.5000 moles of phosphorus triiodide and 2.500 moles of water are used, what is the theoretical yield of hydrogen iodide?
Hint: Don't forget to include the coefficients from the balanced reaction in your consideration.
A) 63.96 g
B) 205.8 g
C) 191.9 g
D) 319.8 g
E) 383.7 g
Diff: 3
Section: 3.6
117) In a chemical equation, AsF3 + CCl4 → AsCl3 + CCl2F2, the theoretical yield of CCl2F2 was calculated to be 1.68 moles. If the percent yield in the reaction was 74.3%, how many grams of CCl2F2 were obtained?
A) 151 grams
B) 167 grams
C) 203 grams
D) 273 grams
E) 303 grams
Diff: 2
Section: 3.7
118) In a chemical equation, AsF3 + C2Cl6 → AsCl3 + C2Cl2F4, the theoretical yield of C2Cl2F4 was calculated to be 1.86 moles. If the percent yield in the reaction was 77.2%, how many grams of C2Cl2F4 were actually obtained?
A) 222 grams
B) 231 grams
C) 245 grams
D) 318 grams
E) 412 grams
Diff: 2
Section: 3.7
119) In a chemical equation, C2H6O + PCl3 → C2H5Cl + H3PO3, when the reaction was carried out, the actual yield of C2H5Cl was calculated as 97.3 % of the theoretical value. If the theoretical yield should have been 2.04 moles, how many grams of C2H5Cl were actually obtained?
A) 123 grams
B) 128 grams
C) 132 grams
D) 135 grams
E) 138 grams
Diff: 1
Section: 3.7
120) In a chemical equation, C3H6O2 + PCl3 → C3H5OCl + H3PO3, when the reaction was carried out, the actual yield of C3H5OCl was calculated as 97.3 % of the theoretical value. If the theoretical yield should have been 1.42 moles, how many grams of C3H5OCl were actually obtained?
A) 115 grams
B) 128 grams
C) 156 grams
D) 193 grams
E) 206 grams
Diff: 1
Section: 3.7
121) You are given the chemical equation: AsF3 + C2Cl6 AsCl3 + C2Cl2F4, if 1.3618 moles of AsF3 are allowed to react with 1.000 moles of C2Cl6, what would be the theoretical yield of C2Cl2F4, in grams?
A) 128.1 grams
B) 134.1 grams
C) 170.9 grams
D) 174.6 grams
E) 185.5 grams
Diff: 2
Section: 3.7
122) Phosphorus tribromide (PBr3, 270.69 g mol-1) and water (18.015 g mol-1) react to form phosphorous acid (H3PO3, 81.996 g mol-1) and hydrogen bromide (80.912 g mol-1). If 0.5000 moles of phosphorus tribromide was reacted with 2.000 moles of water and 98.048 grams of hydrogen bromide were obtained, what was the percent yield from the reaction?
Hint: Don't forget to include the coefficients from the balanced reaction in your consideration.
A) 72.16 %
B) 97.22 %
C) 78.62 %
D) 85.93 %
E) 80.79 %
Diff: 3
Section: 3.7
123) The first step in the Ostwald process for producing nitric acid is
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g).
If the reaction of 150. g of ammonia with 150. g of oxygen gas yields 87. g of nitric oxide (NO), what is the percent yield of this reaction?
A) 100 %
B) 49 %
C) 77 %
D) 33 %
E) 62 %
Diff: 2
Section: 3.7
124) An average atom of uranium (U) is approximately how many times heavier than an atom of phosphorus?
Diff: 1
Section: 3.1
125) A 30.0 g sample of a copper containing compound is found to contain 14.18 g of copper. If the only other element in the sample is chlorine, what is the empirical formula for the compound?
Diff: 3
Section: 3.1
126) A sample of naturally occurring carbon contains 0.800 moles of C. How many atoms are in the sample?
Diff: 1
Section: 3.1
127) The mass of 3.00 moles of phosphorus (P4), is ________ grams.
Diff: 1
Section: 3.1
128) A sample of Na2SO4 weighing 7.10 grams contains ________ moles.
Diff: 2
Section: 3.1
129) Two moles of NH4ClO4 contains ________ moles of nitrogen atoms.
Diff: 1
Section: 3.2
130) There are ________ moles of oxygen atoms in 10 moles of KClO3.
Diff: 1
Section: 3.2
131) There are ________ moles of oxygen atoms in 25.7 g of CaSO4.
Diff: 2
Section: 3.2
132) There are ________ sulfur atoms in 25.6 g of Al2(S2O3)3.
Hint: You can't convert directly from mass of a compound to atoms of an element. Use an intermediary.
Diff: 3
Section: 3.2
133) There are ________ moles of Cl atoms in 65.2 g of CHCl3.
Diff: 2
Section: 3.2
134) There are 7.28 moles of oxygen and ________ moles of iron in a sample of the compound. Fe3O4.
Diff: 1
Section: 3.2
135) A sample of ozone, O3, contains 3.011 × 1012 atoms, which is equivalent to ________ moles of ozone.
Diff: 2
Section: 3.2
136) How many grams of iron are there in a sample of Fe3O4, which has a mass of 8.338 grams?
Diff: 2
Section: 3.2
137) The percent, by weight, of sulfur in (NH4)2SO4 is ________.
Diff: 2
Section: 3.3
138) The percent, by weight, of oxygen in Ca(NO3)2 is ________.
Diff: 2
Section: 3.3
139) How many grams of oxygen are present in 100.0 g Ca(NO3)2 is ________.
Diff: 2
Section: 3.3
140) In the course of determination of a chemical formula, a student obtained the following mole ratios:
C: 1.67 H: 2.67 O: 1.00.
The empirical formula for the compound is ________.
Diff: 1
Section: 3.4
141) In the course of determination of a chemical formula, a student obtained the following mole ratios:
H: 1.67 P: 1.00 O: 2.33.
The empirical formula for the compound is ________.
Diff: 1
Section: 3.4
142) In the course of determination of a chemical formula, a student obtained the following mole ratios:
C: 1.33 H: 2.67 O: 1.00.
The empirical formula for the compound is ________.
Diff: 1
Section: 3.4
143) In the course of determination of a chemical formula, a student obtained the following mole ratios:
C: 1.67 H: 3.33 O: 1.00.
The empirical formula for the compound is ________.
Diff: 1
Section: 3.4
144) A compound contains 46.46% lithium by mass and 53.54 % oxygen by mass. The empirical formula of the compound is ________.
Diff: 2
Section: 3.4
145) The percent composition by mass of a compound is 76.0% C, 12.8% H, and 11.2% O. The molar mass of this compound is found to be 284.5 g/mol. What is the molecular formula of this compound?
Diff: 2
Section: 3.4
146) A 1.375 g sample of mannitol, a sugar found in seaweed, is burned completely in oxygen to give 1.993 g of carbon dioxide and 0.9519 g of water. The empirical formula for mannitol is ________.
Hint: Mannitol contains oxygen as well as carbon and hydrogen.
Diff: 3
Section: 3.4
147) The empirical formula of a compound composed of uranium and fluorine that has a composition of 67.6% uranium and 32.4% fluorine is ________.
Diff: 2
Section: 3.4
148) When the equation is balanced, Al2O3 + C → Al + CO2, the coefficient for CO2 is ________.
Diff: 1
Section: 3.5
149) When the equation is balanced, PCl3(l) + H2O(l) → H3PO3(aq) + HCl(aq), the coefficient of H2O is ________.
Diff: 2
Section: 3.5
150) When the equation is balanced, SF4 + H2O → H2SO3 + HF, the coefficient of HF is ________.
Diff: 2
Section: 3.5
151) When the equation is balanced, CF4 + Cl2 → CCl4 + HCl, the coefficient of HCl is ________.
Diff: 2
Section: 3.5
152) Aluminum hydroxide reacts with nitric acid to form aluminum nitrate and water. If 15.0 g of aluminum hydroxide reacts with excess nitric acid the mass of water that can be formed is ________.
Diff: 2
Section: 3.5
153) For the given process:
TiCl4(g) + 2Mg(l) → 2MgCl2(l) + Ti(s)
If 1 mole of magnesium is reacted with an excess of TiCl4 gas, how many moles of titanium (Ti) can be produced?
Diff: 1
Section: 3.5
154) When Al reacts with Cl2 to form AlCl3, how many moles of Al are required to produce 1.45 moles of AlCl3?
Diff: 2
Section: 3.5
155) Given the chemical equation, 2AsI3 → 2As + 3I2, how many grams of I2 are produced in the reaction, if 3.200 g As are obtained from AsI3?
Diff: 2
Section: 3.5
156) Given the chemical equation, Cl2 + F2 → 2 ClF, how many moles of ClF are produced in the reaction, if 1.50 moles Cl2 are made to react with 1.75 moles F2?
Diff: 1
Section: 3.6
157) Vanadium oxide, V2O5, reacts with calcium according to the chemical equation below. When 10.0 moles of V2O5 are mixed with 10.0 moles of Ca, which is the limiting reagent?
V2O5(s) + 5Ca(l) → 2V(l) + 5CaO(s)
Diff: 1
Section: 3.6
158) When 2.0 moles of V2O5 react with 6.0 mole of calcium, the theoretical yield of vanadium that can be produced is ________ moles, based on the following chemical equation:
V2O5(s) + 5Ca(l) → 2V(l) + 5CaO(s)
Diff: 1
Section: 3.6
159) How many grams of water could be made from 5.0 mol H2 and 3.0 mol O2?
Diff: 2
Section: 3.6
160) Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor:
4NH3 + 5O2 → 4NO + 6H2O
When 20.0 g NH3 and 50.0 g O2 are allowed to react, which is the limiting reagent?
Diff: 2
Section: 3.6
161) Hydrochloric acid can be prepared by the following reaction:
2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s)
How many grams of HCl can be prepared from 2.00 mol H2SO4 and 150 g NaCl?
Hint: Use the limiting reagent when performing your calculations.
Diff: 3
Section: 3.6
162) Calculate the mass of FeS formed when 9.42 g of Fe reacts with 8.50 g of S.
Fe(s) + S(s) → FeS(s)
Hint: Determine the limiting reagent to perform your calculations.
Diff: 3
Section: 3.6
163) Calculate the mass of excess reagent remaining at the end of the reaction in which 90.0 g of SO2 are mixed with 100.0 g of O2.
2SO2 + O2 → 2SO3
Hint: Determine the limiting reagent to perform your calculations.
Diff: 3
Section: 3.6
164) 50.0 grams of NH3 were reacted with 110.0 g of O2 to produce NO and H2O. The theoretical yield of NO, in grams, is
Hint: Write out the equation and balance it, and then determine the limiting reagent to perform your calculations.
A) 206.3 grams.
B) 103.2 grams.
C) 88.1 grams.
D) 123.8 grams.
E) 82.5 grams.
Diff: 3
Section: 3.6
165) Given the chemical equation, CO + 2H2 → CH3OH, if 11.5 g CO react with 11.5 g H2, then the limiting reactant is ________.
Diff: 2
Section: 3.6
166) What is the expected yield of Pb, if 25.0 g C react with 25.0 g of PbO according to the chemical equation, PbO + C → CO + Pb?
Diff: 2
Section: 3.6
167) If 15.3 g of a particular product were obtained in a reaction and the theoretical amount was calculated to be 19.12 g, what was the percent yield of the product?
Diff: 1
Section: 3.7
168) Given the chemical equation, 2Mg + O2 → 2MgO, when 2.2 g Mg react with 3.6 g of O2, 2.7 g MgO were obtained. What is the percent yield in the reaction?
Diff: 2
Section: 3.7
169) When 0.500 moles of boron trichloride react with 1.20 moles hydrogen gas to produce elemental boron and hydrogen chloride gas, the actual yield of elemental boron was 66.4 % of the theoretical yield. The mass of boron obtained was ________.
Diff: 2
Section: 3.7
170) For the given process:
TiCl4(g) + 2Mg(l) → 2MgCl2(l) + Ti(s)
the theoretical yield for titanium (Ti(s)) was calculated to be 305.6 g. The actual amount of titanium recovered was 285.5 g. What was the percent yield for the process?
Diff: 1
Section: 3.7
171) When octane (C8H18) is burned in a particular internal combustion engine, the yield of products (carbon dioxide and water) is 93%. What mass of carbon dioxide will be produced in this engine when 15.0 g of octane is burned with 15.0 g of oxygen gas?
Diff: 1
Section: 3.7
172) In combustion analysis, the compound, CH3NH2, undergoes complete combustion in excess
oxygen to produce a stream of gases. The stream of gases passes through a pre-weighed tube containing anhydrous calcium sulfate and then through a second pre-weighed tube containing sodium hydroxide. The anhydrous calcium sulfate is used to absorb ________.
Diff: 2
Section: On the Cutting Edge 3.1
173) In combustion analysis, the compound, CH3CH3, undergoes complete combustion in excess
oxygen to produce a stream of gases. The stream of gases passes through a pre-weighed tube containing anhydrous calcium sulfate and then through a second pre-weighed tube containing sodium hydroxide. The sodium hydroxide is used to absorb ________.
Diff: 2
Section: On the Cutting Edge 3.1
174) In combustion analysis, a hydrocarbon compound undergoes complete combustion in excess
oxygen to produce a stream of gases. The stream of gases passes through a pre-weighed tube. The increase in mass of these tubes represents the combined mass of CO2 and H2O. The mass of the element ________ in the hydrocarbon can be directly calculated from the mass of CO2.
Diff: 2
Section: On the Cutting Edge 3.1
175) In combustion analysis, a hydrocarbon compound undergoes complete combustion in excess
oxygen to produce a stream of gases. The stream of gases passes through a pre-weighed tube. The increase in mass of these tubes represents the combined mass of CO2 and H2O. The mass of the element ________ in the hydrocarbon can be directly calculated from the mass of H2O.
Diff: 2
Section: On the Cutting Edge 3.1
176) A mole of oxygen, O2, and a mole of phosphorus, P4, do not contain the same number of molecules.
Diff: 1
Section: 3.1
177) A mole of oxygen, O2, and a mole of ozone, O3, contain the same number of molecules.
Diff: 1
Section: 3.1
178) A mole of naturally occurring Ne has a greater mass than a mole of O2.
Diff: 1
Section: 3.1
179) A mole of nitrogen gas, N2, and a mole of carbon dioxide gas, CO2, contain the same number of molecules.
Diff: 1
Section: 3.1
180) A sample of C7H5N3O4 contains 0.255 moles of hydrogen atoms. This same sample also contains 0.500 moles of oxygen atoms.
Diff: 1
Section: 3.2
181) The molecular formula for a substance can never contain fewer atoms of each element than the empirical formula for the same substance.
Diff: 1
Section: 3.4
182) The reaction between 1.12 moles of K and H2O, will produce more than 2.00 moles of KOH.
Diff: 2
Section: 3.5
183) When 1.20 moles C react with 1.20 moles O2, 2.40 moles CO2 are obtained.
Diff: 2
Section: 3.6
184) Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor:
4NH3 + 5O2 → 4NO + 6H2O
When 40.0 g NH3 and 50.0 g O2 are allowed to react, the limiting reagent is O2.
Diff: 2
Section: 3.6
185) In a chemical reaction, the best yield currently obtainable in any process is 99.5%.
Diff: 2
Section: 3.7
186) When the compound, CH3NH2, undergoes complete combustion in excess oxygen in the presence of catalysts, the products are only CO2 and H2O.
Diff: 2
Section: On the Cutting Edge 3.1
187) When the compound, CH3CH2SH, undergoes complete combustion in excess oxygen in the presence of catalysts, the products are only CO2 and H2O.
Diff: 2
Section: On the Cutting Edge 3.1
188) In combustion analysis, if a compound contains oxygen, carbon, and hydrogen, then the mass of oxygen would be determined by subtracting the mass of hydrogen and carbon produced from the total mass of the compound burned.
Diff: 2
Section: On the Cutting Edge 3.1
189) In combustion analysis, all the carbon atoms in CH3(CH2)12CO2H would be converted to carbon dioxide.
Diff: 2
Section: On the Cutting Edge 3.1
190) You have made a bet with a friend. You tell them that you can eat one mole of M&Ms in a year, if you eat one M&M every second for three hours a day, 7 days a week, for 365 days in that year. Can you win the bet? Calculate how many years it would really take to eat one mole of M&Ms.
Hint: Use the amount of time eating M&Ms to calculate the number eaten.
Diff: 3
Section: 3.1
191) A 5.21 g sample of an unknown element is found to contain 4.937 × 1022 atoms. Based on this information, what is the unknown element?
Hint: Determine the molar mass of the unknown element.
Diff: 3
Section: 3.1
192) How many carbon atoms are there in 10 lbs of sugar, C12H22O11?
Hint: 1 lb = 453.592 g.
Diff: 3
Section: 3.2
193) The imaginary compound calcium nortonate, CaNtO4, is 54.698% nortonium, Nt, by weight. Calculate the percent, by weight, of sodium in sodium nortonate to four significant digits.
Hint: Don't forgot the rules about how to combine ions to create neutral ionic substances, to predict the formula of sodium nortonate.
Diff: 3
Section: 3.4
194) The imaginary compound calcium nortonate, CaNtO4, is 54.698% nortonium, Nt, by weight. Calculate the percent, by weight, of potassium in potassium nortonate to four significant digits.
Hint: Don't forgot the rules about how to combine ions to create neutral ionic substances, to predict the formula of potassium nortonate.
Diff: 3
Section: 3.4
195) The compound, X2O3, contains element X and oxygen combined in a ratio of 2.167 grams of X to 1.000 gram of oxygen. Another compound containing element X and oxygen gave a different analysis and properties—a 2.500 g sample of this second compound contains 1.300 grams of element X. This works out to a different ratio. What is the formula for the second compound?
Hint: The mass of each X atom will be the same and this can be used to find molar ratios needed in the second compound.
Diff: 3
Section: 3.4
196) A compound which has the formula, Z2O3, contains element Z and oxygen combined in a ratio of 2.123 grams of element Z to 1.000 gram of oxygen. A 2.500 gram sample of a different compound containing the same two elements was analyzed and found to contain 1.400 grams of element Z. What is the formula for this second compound?
Hint: The mass of each Z atom will be the same and this can be used to find molar ratios needed in the second compound.
Diff: 3
Section: 3.4
197) In a quantitative study, 4.624 grams of a compound containing carbon, hydrogen, and oxygen yielded 7.210 g of CO2 and 2.656 g of H2O in a combustion apparatus. Determine the empirical formula of the compound.
Hint: Use a multiplier that gives you molar ratios that are precise to 2 decimals to avoid rounding errors.
A) C9H16O8
B) C10H18O9
C) C11H20O10
D) C12H22O11
E) C13H24O12
Diff: 3
Section: 3.5
198) A 5.000 g sample of a mixture which contains MgCO3 and sand (SiO2) was heated for 2.30 hours until no further reaction occurred. It was then cooled and the residue, which contained MgO and unchanged sand, was weighed. The reaction is:
MgCO3(s) → MgO(s) + CO2(g).
The residue remaining (minus the CO2, which was completely driven off) weighed 3.397 grams. Calculate the percent, by weight, of MgCO3 in the original sample.
Hint: Use moles as an intermediary when relating different units together.
A) 22.59 %
B) 50.00 %
C) 61.42 %
D) 90.40 %
E) 95.01 %
Diff: 3
Section: 3.5
199) A mixture containing silver nitrate (AgNO3) and potassium nitrate (KNO3), weighing 5.000 grams, was dissolved in water and treated with an excess of potassium chloride solution. A quantitative reaction, the reaction of silver nitrate with potassium chloride occurred, and 1.582 grams of a white precipitate, silver chloride, was obtained. Determine the percent, by weight, of silver nitrate in the original mixture.
Hint: Use moles as an intermediary when relating different units together.
A) 32.55 %
B) 37.50 %
C) 42.50 %
D) 62.50 %
E) 62.74 %
Diff: 3
Section: 3.5
200) Ores can be composed of many different components, including lead sulfide (PbS). The amount of lead sulfide in an ore can be determined through the reaction:
2 PbS(s) + 6 HNO3(aq) + K2Cr2O7(aq)
2 PbCrO4(s) + 2 S(s) + 4 NO2(g) + 2 KNO3(aq) + 3 H2O
If a 6.053 g sample of ore that was treated with excess nitric acid and potassium dichromate yielded 6.094 g of PbCrO4, calculate the percent, by weight, of lead sulfide in the ore.
Hint: Convert the masses to moles and calculate as you would for other stoichiometric calculations.
A) 68.22%
B) 74.53%
C) 77.22%
D) 79.16%
E) 83.11%
Diff: 3
Section: 3.5
201) The mineral dolomite has the formula, CaCO3∙MgCO3. When dolomite is heated for an extended period of time at elevated temperatures, it decomposes into carbon dioxide and a mixture of calcium oxide and magnesium oxide. If a 5.424 g sample of dolomite is heated for several hours at 950 °C, what should be the mass of the mixed oxides produced?
Hint: Write out the balanced chemical equation before proceeding.
Diff: 3
Section: 3.5
202) A laboratory unknown is a mixture of calcium carbonate and magnesium carbonate. These substances decompose to form calcium oxide and magnesium oxide, respectively, when heated at a high temperature for a prolonged period. A sample of this laboratory unknown mixture with a mass of 5.424 grams was decomposed by heating for several hours at 950 °C. The oxide residue produced had a mass of 2.791 grams. From this laboratory data calculate the percent, by weight, of calcium carbonate in the unknown.
Hint: Use the balanced chemical equations to find the molar ratios of the carbonates to the oxides, then create two algebraic equations — one for the carbonates and the other for the products. This will give you two equations with two unknowns, which can be solved to determine the number of moles of one of your substances.
Diff: 3
Section: 3.5
203) A chemistry graduate student needs to make a "starting material" for a series of experiments. The student is going to follow the balanced equation shown below:
2Mo(CO)6(s) + 4 (CH3)3CCO2H (s) → Mo2(O2C(CH3)3)4(s) + 12CO(g) + 2H2(g)
If the student uses 5.00 grams of (CH3)3CCO2H, how many grams of Mo(CO)6 will the student need to react completely with the (CH3)3CCO2H?
Diff: 2
Section: 3.5
204) A chemistry graduate student needs to make a "starting material" for a series of experiments. The student is going to follow the balanced equation shown below:
2Mo(CO)6(s) + 4 (CH3)3CCO2H (s) → Mo2(O2C(CH3)3)4(s) + 12CO(g) + 2H2(g)
If the student uses 5.00 grams of (CH3)3CCO2H, how many grams of Mo2(O2C(CH3)3)4 will be made?
Diff: 2
Section: 3.5
205) One way of making sodium chloride is according to the following unbalanced equation:
Na(s) + Cl2(g) → NaCl(s)
In a reaction, a student mixed 2.00 grams of Na(s) with 4.13 grams of Cl2(g), making 5.00 grams of NaCl. Balance the equation and determine which, if any, of the starting materials was in excess?
Diff: 2
Section: 3.6
206) The reactants and products in a reaction are shown below in the unbalanced chemical equation:
P4O10 + Ca(OH)2 → Ca3(PO4)2 + H2O
If you start with 2.40 moles of P4O10 and 3.30 moles of Ca(OH)2 and all of the limiting reactant is consumed,
Hint: You must balance the equation first.
A) 1.10 moles of Ca3(PO4)2 are produced.
B) 0.90 moles of Ca(OH)2 are left over.
C) 2.40 moles of Ca3(PO4)2 are produced.
D) 1.65 moles of Ca3(PO4)2 are produced.
E) 0.85 moles of P4O10 are consumed.
Diff: 3
Section: 3.6
207) A mixture, weighing 6.000 grams, contains 41.50 % Na3PO4 and 58.50 % BaCl2 of the mixture, by weight. When dissolved in water, a precipitate forms via the double displacement reaction:
BaCl2(aq) + Na3PO4(aq) NaCl(aq) + Ba3(PO4)2(s)
If the reaction is quantitative, the yield of the solid barium phosphate should be
Hint: You must balance the equation first then determine the limiting reagent.
A) 3.382 grams.
B) 3.357 grams.
C) 4.571 grams.
D) 3.510 grams.
E) 1.983 grams.
Diff: 3
Section: 3.6
208) Liquid hexane, C6H14, burns in oxygen gas to yield carbon dioxide and water. What is the minimum mass of oxygen required for the complete reaction of 10.0 mL of hexane?
[Given: density of hexane = 0.660 g/mL]
Hint: You need to use the information given to find the starting mass of hexane.
A) 3.71 g
B) 2.45 g
C) 23.3 g
D) 46.6 g
E) 35.3 g
Diff: 3
Section: 3.6
209) A student is working on a research project. The instructions in the book on how to prepare Ni(NH3)6Cl2 say that the percent yield in this preparation is 75.5 %. If the limiting reagent for the preparation is NiCl2∙6H2O, and the student needs 50.0 grams of the Ni(NH3)6Cl2 product for the next step of the research project, how many grams of the NiCl2·6H2O should the student start with to prepare the desired quantity of the Ni(NH3)6Cl2?
Hint: Since you cannot write out a balanced equation from the information given, use the ratio of Ni in the reactant and product to figure out the stoichiometric ratio of the substances.
Diff: 3
Section: 3.7
210) A laboratory manual gave precise instructions for carrying out the reaction described by the chemical equation, C2H6O + O2 → C2H4O2 + H2O. It stated that if you bubble oxygen gas through a solution containing the C2H6O for 24 hours, the yield of C2H4O2 would be 7.50% of the theoretical amount. If you need to produce 700 grams of the C2H4O2 in a single batch, how many kg of C2H6O should you begin with?
Hint: You need to first determine how much product you need to aim for to accommodate an actual yield of only 7.5%.
Diff: 3
Section: 3.7
211) The Hall process for the production of aluminum involves the reaction of aluminum oxide with elemental carbon to give aluminum metal and carbon monoxide. If the yield of this reaction is 82% and aluminum ore is 71% by mass aluminum oxide, what mass of aluminum ore must be mined in order to produce 1.0 × 103 kg (1 metric ton) of aluminum metal by the Hall process?
Hint: You need to first determine how product you need to aim for to accommodate an actual yield of 82%, and then account for the fact that the ore is only 71% aluminum oxide.
Diff: 3
Section: 3.7
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