CHAPTER THREE STOICHIOMETRY - … · 29 CHAPTER THREE STOICHIOMETRY Questions 18. The two major isotopes of boron are 10B and 11B.The listed mass of 10.81 is the average mass of a

29

CHAPTER THREE

STOICHIOMETRY

Questions

18. The two major isotopes of boron are B and B. The listed mass of 10.81 is the average mass of a10 11

very large number of boron atoms.

19. The molecular formula tells us the actual number of atoms of each element in a molecule (or formulaunit) of a compound. The empirical formula tells only the simplest whole number ratio of atoms ofeach element in a molecule. The molecular formula is a whole number multiple of the empiricalformula. If that multiplier is one, the molecular and empirical formulas are the same. For example,

2both the molecular and empirical formulas of water are H O. They are the same. For hydrogen

2 2peroxide, the empirical formula is OH; the molecular formula is H O .

4 2 220. Side reactions may occur. For example, in the combustion of CH (methane) to CO and H O, someCO is also formed. Also, some reactions only go part way to completion and reach a state ofequilibrium where both reactants and products are present (see Ch. 13).

Exercises

Atomic Masses and the Mass Spectrometer

21. A = atomic mass = 0.7899(23.9850 amu) + 0.1000(24.9858 amu) + 0.1101(25.9826 amu)

A = 18.95 amu + 2.499 amu + 2.861 amu = 24.31 amu

22. A = 0.0140(203.973) + 0.2410(205.9745) + 0.2210(206.9759) + 0.5240(207.9766)

A = 2.86 + 49.64 + 45.74 + 109.0 = 207.2 amu; From the periodic table, the element is Pb.

23. Let x = % of Eu and y = % of Eu, then x + y = 100 and y = 100 - x.151 153

151.96 =

15196 = 150.9196 x + 15292.09 - 152.9209 x, -96 = -2.0013 x

x = 48%; 48% Eu and 100 - 48 = 52% Eu151 153

CHAPTER 3 STOICHIOMETRY30

24. Let A = mass of Re: 186.207 = 0.6260(186.956) + 0.3740(A), 186.207 - 117.0 = 0.3740(A)185

A = = 185 amu (A = 184.95 amu without rounding to proper significant figures.)

25. There are three peaks in the mass spectrum, each 2 mass units apart. This is consistent with two

2isotopes, differing in mass by two mass units. The peak at 157.84 corresponds to a Br moleculecomposed of two atoms of the lighter isotope. This isotope has mass equal to 157.84/2 or 78.92. Thiscorresponds to Br. The second isotope is Br with mass equal to 161.84/2 = 80.92. The peaks in 79 81

2 2the mass spectrum correspond to Br , Br Br, and Br in order of increasing mass. The intensities79 79 81 81

of the highest and lowest mass tell us the two isotopes are present in about equal abundance. Theactual abundance is 50.69% Br and 49.31% Br. The calculation of the abundance from the mass79 81

spectrum is beyond the scope of this text.

26. GaAs can be either GaAs or GaAs. The mass spectrum for GaAs will have 2 peaks at 14469 71

(= 69 + 75) and 146 (= 71 + 75) with intensities in the ratio of 60:40 or 3:2.

2 2 2 2 2 2 2Ga As can be Ga As , Ga GaAs , or Ga As . The mass spectrum will have 3 peaks at 288, 290,69 69 71 71

and 292 with intensities in the ratio of 36:48:16 or 9:12:4. We get this ratio from the followingprobability table:

Ga (0.60) Ga (0.40)69 71

Ga (0.60) 0.36 0.2469

Ga (0.40) 0.24 0.1671

CHAPTER 3 STOICHIOMETRY 31

Moles and Molar Masses

27. When more than one conversion factor is necessary to determine the answer, we will usually put allthe conversion factors into one calculation instead of determining intermediate answers. This methodreduces round-off error and is a time saver.

500. atoms Fe × = 4.64 × 10 g Fe-20

28. 500.0 g Fe × = 8.953 mol Fe

8.953 mol Fe × = 5.391 × 10 atoms Fe24

29. 1.00 carat × = 1.00 × 10 atoms C22

30. 5.0 × 10 atoms C × = 8.3 × 10 mol C21 -3

8.3 × 10 mol C × = 0.10 g C-3

2 331. Al O : 2(26.98) + 3(16.00) = 101.96 g/mol

3 6Na AlF : 3(22.99) + 1(26.98) + 6(19.00) = 209.95 g/mol

2 332. HFC - 134a, CH FCF : 2(12.01) + 2(1.008) + 4(19.00) = 102.04 g/mol

3HCFC-124, CHClFCF : 2(12.01) + 1(1.008) + 1(35.45) + 4(19.00) = 136.48 g/mol

333. a. The formula is NH . 14.01 g/mol + 3(1.008 g/mol) = 17.03 g/mol

2 4b. The formula is N H . 2(14.01) + 4(1.008) = 32.05 g/mol

4 2 2 7c. (NH ) Cr O : 2(14.01) + 8(1.008) + 2(52.00) + 7(16.00) = 252.08 g/mol

4 634. a. The formula is P O . 4(30.97 g/mol) + 6(16.00 g/mol) = 219.88 g/mol

3 4 2b. Ca (PO ) : 3(40.08) + 2(30.97) + 8(16.00) = 310.18 g/mol

2 4c. Na HPO : 2(22.99) + 1(1.008) + 1(30.97) + 4(16.00) = 141.96 g/mol

3 335. a. 1.00 g NH × = 0.0587 mol NH


2 4 2 4b. 1.00 g N H × = 0.0312 mol N H

4 2 2 7 4 2 2 7c. 1.00 g (NH ) Cr O × = 3.97 × 10 mol (NH ) Cr O-3

4 6 4 636. a. 1.00 g P O × = 4.55 × 10 mol P O-3

3 4 2 3 4 2b. 1.00 g Ca (PO ) × = 3.22 × 10 mol Ca (PO )-3

2 4 2 4c. 1.00 g Na HPO × = 7.04 × 10 mol Na HPO-3

3 337. a. 5.00 mol NH × = 85.2 g NH

2 4 2 4b. 5.00 mol N H × = 160. g N H

4 2 2 7 4 2 2 7c. 5.00 mol (NH ) Cr O × = 1260 g (NH ) Cr O

4 6 4 638. a. 5.00 mol P O × = 1.10 × 10 g P O3

3 4 2 3 4 2b. 5.00 mol Ca (PO ) × = 1.55 × 10 g Ca (PO )3

2 4 2 4c. 5.00 mol Na HPO × = 7.10 × 10 g Na HPO2

39. Chemical formulas give atom ratios as well as mol ratios.

3a. 5.00 mol NH × = 70.1 g N

2 4b. 5.00 mol N H × = 140. g N

4 2 2 7c. 5.00 mol (NH ) Cr O × = 140. g N

4 640. a. 5.00 mol P O × = 619 g P


3 4 2b. 5.00 mol Ca (PO ) × = 310. g P

2 4c. 5.00 mol Na HPO × = 155 g P

3 341. a. 1.00 g NH × = 3.54 × 10 molecules NH22

2 4 2 4b. 1.00 g N H × = 1.88 × 10 molecules N H22

4 2 2 7c. 1.00 g (NH ) Cr O ×

4 2 2 7= 2.39 × 10 formula units (NH ) Cr O21

4 6 4 642. a. 1.00 g P O × = 2.74 × 10 molecules P O21

3 4 2b. 1.00 g Ca (PO ) ×

3 4 2 = 1.94 × 10 formula units Ca (PO )21

2 4c. 1.00 g Na HPO ×

2 4 = 4.24 × 10 formula units Na HPO21

43. Using answers from Exercise 41:

3a. 3.54 × 10 molecules NH × = 3.54 × 10 atoms N22 22

2 4b. 1.88 × 10 molecules N H × = 3.76 × 10 atoms N22 22

4 2 2 7c. 2.39 × 10 formula units (NH ) Cr O × = 4.78 × 10 atoms N21 21

44. Using answers from Exercise 42:

4 6a. 2.74 × 10 molecules P O × = 1.10 × 10 atoms P21 22


3 4 2b. 1.94 × 10 formula units Ca (PO ) × = 3.88 × 10 atoms P21 21

2 4c. 4.24 × 10 formula units Na HPO × = 4.24 × 10 atoms P21 21

6 8 645. Molar mass of C H O = 6(12.01) + 8(1.008) + 6(16.00) = 176.12 g/mol

500.0 mg × = 2.839 × 10 mol-3

2.839 × 10 mol × = 1.710 × 10 molecules-3 21

46. a. 9(12.01) + 8(1.008) + 4(16.00) = 180.15 g/mol

b. 500. mg × = 2.78 × 10 mol-3

2.78 × 10 mol × = 1.67 × 10 molecules-3 21

47. a. 2.49 × 10 molecules CO × = 4.13 × 10 mol CO20 -4

4 4b. 15.0 g CuSO × = 9.40 ×10 mol CuSO-2

2 4 2 4c. 100 molecules H SO × = 1.661 × 10 mol H SO-22

2 2d. 6.210 mg K O × = 6.592 × 10 mol K O-5

2 3 2 348. a. 150.0 g Fe O × = 0.9393 mol Fe O

2 2b. 10.0 mg NO × = 2.17 × 10 mol NO-4

3 3c. 1.5 × 10 molecules BF × = 2.5 × 10 mol BF16 -8

2 249. a. 1.27 mmol CO × = 5.59 × 10 g CO-2

3 3b. 2.00 × 10 molecules NCl × = 4.00 g NCl22


4 2 3 4 2 3c. 0.00451 mol (NH ) CO × = 0.433 g (NH ) CO

2 2d. 1 molecule N × = 4.653 × 10 g N-23

4 4e. 62.7 mol CuSO × = 1.00 × 10 g CuSO4

50. a. A chemical formula gives atom ratios as well as mole ratios. We will use both ideas to showhow these conversion factors can be used.

2 5 2Molar mass of C H O N = 2(12.01) + 5(1.008) + 2(16.00) + 14.0l = 75.07 g/mol

2 5 25.00 g C H O N ×

×

3 2b. Molar mass of Mg N = 3(24.31) + 2(14.01) = 100.95 g/mol

3 2 5.00 g Mg N ×

= 5.97 × 10 atoms N 22

3 2c. Molar mass of Ca(NO ) = 40.08 + 2(14.01) + 6(16.00) = 164.10 g/mol

3 25.00 g Ca(NO ) ×

= 3.67 × 10 atoms N22

2 4d. Molar mass of N O = 2(14.01) + 4(16.00) = 92.02 g/mol

2 45.00 g N O × = 6.54 × 10 atoms N22

51. a. 14 mol C + 18 mol H + 2 mol N + 5 mol O

= 294.30 g/mol

b. 10.0 g aspartame × = 3.40 × 10 mol-2


c. 1.56 mol × = 459 g

d. 5.0 mg × = 1.0 × 10 molecules19

e. The chemical formula tells us that 1 molecule of aspartame contains two atoms of N. Thechemical formula also says that 1 mol of aspartame contains two mol of N.

1.2 g aspartame ×

= 4.9 × 10 atoms of nitrogen21

f. 1.0 × 10 molecules × = 4.9 × 10 g or 490 fg9 -13

g. 1 molecule aspartame × = 4.887 × 10 g-22

52. a. 2(12.01) + 3(1.008) + 3(35.45) + 2(16.00) = 165.39 g/mol

b. 500.0 g × = 3.023 mol c. 2.0 × 10 mol × = 3.3 g-2

2 3 3 2d. 5.0 g C H Cl O ×

= 5.5 × 10 atoms of chlorine22

e. 1.0 g Cl × = 1.6 g chloral hydrate

f. 500 molecules × = 1.373 × 10 g-19

Percent Composition

2 3 753. In 1 mole of YBa Cu O , there are 1 mole of Y, 2 moles of Ba, 3 moles of Cu and 7 moles of O.

Molar mass = 1 mol Y + 2 mol Ba

+ 3 mol Cu + 7 mol O


Molar mass = 88.91 + 274.6 + 190.65 + 112.00 = 666.2 g/mol

%Y = × 100 = 13.35% Y; %Ba = × 100 = 41.22% Ba

%Cu = × 100 = 28.62% Cu; %O = × 100 = 16.81% O

3 4 254. a. C H O : Molar mass = 3(12.01) + 4(1.008) + 2(16.00) = 36.03 + 4.032 + 32.00 = 72.06 g/mol

%C = × 100 = 50.00% C; %H = × 100 = 5.595% H

%O = 100.00 - (50.00 + 5.595) = 44.41% O or %O = × 100 = 44.41% O

4 6 2b. C H O : Molar mass = 4(12.01) + 6(1.008) + 2(16.00) = 48.04 + 6.048 + 32.00 = 86.09 g/mol

%C = × 100 = 55.80% C; %H = × 100 = 7.025% H

%O = 100.00 - (55.80 + 7.025) = 37.18% O

3 3c. C H N: Molar mass = 3(12.01) + 3(1.008) + 1(14.01) = 36.03 + 3.024 + 14.01 = 53.06 g/mol

%C = × 100 = 67.90% C; %H = × 100 = 5.699% H

%N = × 100 = 26.40% N or %N = 100.00 - (67.90 + 5.699) = 26.40% N

55. a. NO: %N = × 100 = 46.68% N

2b. NO : %N = × 100 = 30.45% N

2 4c. N O : %N = × 100 = 30.45% N

2d. N O: %N = × 100 = 63.65% N

2 2 4 2The order from lowest to highest mass percentage of nitrogen is: NO = N O < NO < N O.

8 10 4 256. C H N O : molar mass = 8(12.01) + 10(1.008) + 4(14.0l) + 2(16.00) = 194.20 g/mol

%C = × 100 = × 100 = 49.47% C


12 22 11C H O : molar mass = 12(12.01) + 22(1.008) + 11(16.00) = 342.30 g/mol

%C = × 100 = 42.10% C

2 5C H OH: molar mass = 2(12.01) + 6(1.008) + 1(16.00) = 46.07 g/mol

%C = × 100 = 52.14% C

12 22 11The order from lowest to highest mass percentage of carbon is: sucrose (C H O ) < caffeine

8 10 4 2 2 5 (C H N O ) < ethanol (C H OH)

57. There are many valid methods to solve this problem. We will assume 100.00 g of compound; thendetermine from the information in the problem how many mol of compound equals 100.00 g ofcompound. From this information, we can determine the mass of one mol of compound (the molarmass) by setting up a ratio. Assuming 100.00 g cyanocobalamin:

mol cyanocobalamin = 4.34 g Co ×

= 7.36 × 10 mol cyanocobalamin-2

, x = molar mass = 1360 g/mol

58. There are 0.390 g Cu for every 100.00 g of fungal laccase. Assuming 100.00 g fungal laccase:

mol fungal laccase = 0.390 g Cu × = 1.53 × 10 mol -3

, x = molar mass = 6.54 × 10 g/mol4

Empirical and Molecular Formulas

259. a. Molar mass of CH O = 1 mol C + 2 mol H

+ 1 mol O = 30.03 g/mol

%C = × 100 = 39.99% C; %H = × 100 = 6.713% H


%O = × 100 = 53.28% O or %O = 100.00 - (39.99 + 6.713) = 53.30%

6 12 6b. Molar Mass of C H O = 6(12.01) + 12(1.008) + 6(16.00) = 180.16 g/mol

%C = × 100 = 40.00%; %H = × 100 = 6.714%

%O = 100.00 - (40.00 + 6.714) = 53.29%

2 3 2c. Molar mass of HC H O = 2(12.01) + 4(1.008) + 2(16.00) = 60.05 g/mol

%C = × 100 = 40.00%; %H = × 100 = 6.714%

%O = 100.00 - (40.00 + 6.714) = 53.29%

260. All three compounds have the same empirical formula, CH O, and different molecular formulas. Thecomposition of all three in mass percent is also the same (within rounding differences). Therefore,elemental analysis will give us only the empirical formula.

2 461. a. The molecular formula is N O . The smallest whole number ratio of the atoms (the empirical

2formula) is NO .

3 6 2b. Molecular formula: C H ; empirical formula = CH

4 10 2 5c. Molecular formula: P O ; empirical formula = P O

6 12 6 2d. Molecular formula: C H O ; empirical formula = CH O

62. a. SNH: Empirical formula mass = 32.07 + 14.01 + 1.008 = 47.09 g

4 4 4 4 = 4.000; So the molecular formula is (SNH) or S N H .

2b. NPCl : Empirical formula mass = 14.01 + 30.97 + 2(35.45) = 115.88 g/mol

2 3 3 3 6 = 3.0000; Molecular formula is (NPCl ) or N P Cl .

4 4c. CoC O : 58.93 + 4(12.01) + 4(16.00) = 170.97 g/mol

2 8 8 = 2.0000; Molecular formula: Co C O

4 4d. SN: 32.07 + 14.01 = 46.08 g/mol; = 4.000; Molecular formula: S N


63. Out of 100.0 g of the pigment, there are:

59.9 g Ti × = 1.25 mol Ti; 40.1 g O × = 2.51 mol O

2Empirical formula = TiO since mol O to mol Ti are in a 2:1 mol ratio (2.51/1.25 = 2.01).

64. Out of 100.00 g of adrenaline, there are:

56.79 g C × = 4.729 mol C; 6.56 g H × = 6.51 mol H

28.37 g O × = 1.773 mol O; 8.28 g N × = 0.591 mol N

Dividing each mol value by the smallest number:

= 8.00; = 11.0; = 3.00; = 1.00

8 11 3This gives adrenaline an empirical formula of C H O N.

x y65. Compound I: mass O = 0.6498 g Hg O - 0.6018 g Hg = 0.0480 g O

0.6018 g Hg × = 3.000 × 10 mol Hg-3

0.0480 g O × = 3.00 × 10 mol O-3

The mol ratio between Hg and O is 1:1, so the empirical formula of compound I is HgO.

x yCompound II: mass Hg = 0.4172 g Hg O - 0.016 g O = 0.401 g Hg

0.401 g Hg × = 2.00 × 10 mol Hg; 0.016 g O × = 1.0 × 10 mol O-3 -3

2The mol ratio between Hg and O is 2:1, so the empirical formula is Hg O.

66. 1.121 g N × = 8.001 × 10 mol N; 0.161 g H × = 1.60 × 10 mol H-2 -1

0.480 g C × = 4.00 × 10 mol C; 0.640 g O × = 4.00 × 10 mol O-2 -2

Dividing all mol values by the smallest number:

= 2.00; = 4.00; = 1.00

2 4Empirical formula = N H CO

67. Out of 100.0 g compound: 30.4 g N × = 2.17 mol N


%O = 100.0 - 30.4 = 69.6% O; 69.6 g O × = 4.35 mol O

2 = 1.00; = 2.00; Empirical formula is NO .

2The empirical formula mass of NO . 14 + 2(16) = 46 g/mol.

2 4 = 2.0; Therefore, the molecular formula is N O .

68. Out of 100.0 g, there are:

69.6 g S × = 2.17 mol S; 30.4 g N × = 2.17 mol N

Empirical formula is SN since mol values are in a 1:1 mol ratio.

4 4The empirical formula mass of SN is ~ 46 g. Since = 4.0, the molecular formula is S N .

69. Assuming 100.00 g of compound (mass hydrogen = 100.00 g - 49.31 g C - 43.79 g O = 6.90 g H):

49.31 g C × = 4.106 mol C; 6.90 g H × = 6.85 mol H

43.79 g O × = 2.737 mol O

Dividing all mole values by 2.737 gives:

= 1.500; = 2.50; = 1.000

3 5 2Since a whole number ratio is required, the empirical formula is C H O .

The empirical formula mass is: 3(12.01) + 5(1.008) +2(16.00) = 73.07 g/mol

3 5 2 2 6 10 4 = 1.999; molecular formula = (C H O ) = C H O

70. Assuming 100.00 g of compound (mass oxygen = 100.00 g - 41.39 g C - 3.47 g H = 55.14 g O):

41.39 g C × = 3.446 mol C; 3.47 g H × = 3.44 mol H

55.14 g O × = 3.446 mol O

All are the same mol values so the empirical formula is CHO. The empirical formula mass is 12.01 + 1.008 + 16.00 = 29.02 g/mol.


molar mass = = 116 g/mol

4 4 4 4 = 4.00; molecular formula = (CHO) = C H O

71. When combustion data are given, it is assumed that all the carbon in the compound ends up as carbon

2 2in CO and all the hydrogen in the compound ends up as hydrogen in H O. In the sample of propanecombusted, the moles of C and H are:

2mol C = 2.641 g CO × = 0.06001 mol C

2mol H = 1.442 g H O × = 0.1600 mol H

= 2.666

3 8Multiplying this ratio by three gives the empirical formula of C H .

72. This compound contains nitrogen, and one way to determine the amount of nitrogen in the compound

2is to calculate composition by mass percent. We assume that all of the carbon in 33.5 mg CO came

2from the 35.0 mg of compound and all of the hydrogen in 41.1 mg H O came from the 35.0 mg ofcompound.

23.35 × 10 g CO × = 9.14 × 10 g C-2 -3

%C = × 100 = 26.1% C

24.11 × 10 g H O × = 4.60 × 10 g H-2 -3

%H = × 100 = 13.1% H

The mass percent of nitrogen is obtained by difference:

%N = 100.0 - (26.1 + 13.1) = 60.8% N

Now perform the empirical formula determination by first assuming 100.0 g of compound. Out of100.0 g of compound, there are:

26.1 g C × = 2.17 mol C; 13.1 g H × = 13.0 mol H


60.8 g N × = 4.34 mol N

Dividing all mol values by 2.17 gives: = 5.99; = 2.00

6 2The empirical formula is CH N .

73. The combustion data allow determination of the amount of hydrogen in cumene. One way to determinethe amount of carbon in cumene is to determine the mass percent of hydrogen in the compound fromthe data in the problem; then determine the mass percent of carbon by difference (100.0 - mass %H= mass %C).

242.8 mg H O × = 4.79 mg H

%H = × 100 = 10.1% H; %C = 100.0 - 10.1 = 89.9% C

Now solve this empirical formula problem. Out of 100.0 g cumene, we have:

89.9 g C × = 7.49 mol C; 10.1 g H × = 10.0 mol H

3 4 = 1.34 . , i.e., mol H to mol C are in a 4:3 ratio. Empirical formula = C H

Empirical formula mass . 3(12) + 4(1) = 40 g/mol

3 4 3 9 12The molecular formula is (C H ) or C H since the molar mass will be between 115 and 125 g/mol(molar mass . 3 × 40 g/mol = 120 g/mol).

74. First, we will determine composition by mass percent:

216.01 mg CO × = 4.369 mg C

%C = × 100 = 40.91% C

24.37 mg H O × = 0.489 mg H

%H = × 100 = 4.58% H; %O = 100.00 - (40.91 + 4.58) = 54.51% O

So, in 100.00 g of the compound, we have:

40.91 g C × = 3.406 mol C; 4.58 g H × = 4.54 mol H


54.51 g O × = 3.407 mol O

3 4 3Dividing by the smallest number: = 1.33 = ; the empirical formula is C H O .

3 4 3The empirical formula mass of C H O is . 3(12) + 4(1) + 3(16) = 88 g.

6 8 6Since = 2.0, then the molecular formula is C H O .

Balancing Chemical Equations

75. When balancing reactions, start with elements that appear in only one of the reactants and one of the products, then go on to balance the remaining elements.

2 2 3 2 2 3a. Fe + O ÷ Fe O . Balancing Fe first, then O, gives: 2 Fe + 3/2 O ÷ Fe O . The bestbalanced equation contains the smallest whole numbers. To convert to whole numbers, multiply

2 2 3each coefficient by two, which gives: 4 Fe(s) + 3 O (g) ÷ 2 Fe O (s)

2 2 2b. Ca + H O ÷ Ca(OH) + H ; Calcium is already balanced, so concentrate on oxygen next.

2 2 2 Balancing O gives: Ca(s) + 2 H O(l) ÷ Ca(OH) (aq) + H (g). The equation is balanced.Note: Hydrogen is the most difficult element to balance since it appears in both products. It isgenerally easiest to save these atoms for last when balancing an equation.

2 2 4 4 2c. Ba(OH) + H SO ÷ BaSO + H O; Ba and S are already balanced. There are 6 O atoms

2on the reactant side and, in order to get 6 O atoms on the product side, we will need 2 H O

2 2 4 4 2molecules. The balanced equation is: Ba(OH) (aq) + H SO (aq)÷ BaSO (s) + 2 H O(l).

6 12 6 2 2 276. a. C H O (s) + O (g) ÷ CO (g) + H O(g)

6 12 6 2 2 2Balance C atoms: C H O + O ÷ 6 CO + H O

6 12 6 2 2 2Balance H atoms: C H O + O ÷ 6 CO + 6 H O

6 12 6 2 2 2Lastly, balance O atoms: C H O (s) + 6 O (g) ÷ 6 CO (g) + 6 H O(g)

The equation is balanced.

2 3 3 2b. Fe S (s) + HCl(g) ÷ FeCl (s) + H S(g)

2 3 3 2Balance Fe atoms: Fe S + HCl ÷ 2 FeCl + H S

2 3 3 2Balance S atoms: Fe S + HCl ÷ 2 FeCl + 3 H S

There are 6 H and 6 Cl on right, so balance with 6 HCl on left:


2 3 3 2Fe S (s) + 6 HCl(g) ÷ 2 FeCl (s) + 3 H S(g). Equation is balanced.

2 3 2 4c. CS (l) + NH (g) ÷ H S(g) + NH SCN(s)

C and S balanced; balance N:

2 3 2 4CS + 2 NH ÷ H S + NH SCN

2 3 2 4H is also balanced. So: CS (l) + 2 NH (g) ÷ H S(g) + NH SCN(s)

3 3 277. a. Cu(s) + 2 AgNO (aq) ÷ 2 Ag(s) + Cu(NO ) (aq)

2 2b. Zn(s) + 2 HCl(aq) ÷ ZnCl (aq) + H (g)

2 3 2 2c. Au S (s) + 3 H (g) ÷ 2 Au(s) + 3 H S(g)

2 3 4 2 3 4 278. a. 3 Ca(OH) (aq) + 2 H PO (aq) ÷ 6 H O(l) + Ca (PO ) (s)

3 3 2b. Al(OH) (s) + 3 HCl(aq) ÷ AlCl (aq) + 3 H O(l)

3 2 4 2 4 3c. 2 AgNO (aq) + H SO (aq) ÷ Ag SO (s) + 2 HNO (aq)

6 6 2 2 279. a. The formulas of the reactants and products are C H (l) + O (g) ÷ CO (g) + H O(g).

6 6To balance this combustion reaction, notice that all of the carbon in C H has to end up as carbon

2 6 6 2in CO and all of the hydrogen in C H has to end up as hydrogen in H O. To balance C and H,

2 2 6 6we need 6 CO molecules and 3 H O molecules for every 1 molecule of C H . We do oxygen last.

2 2 2 Since we have 15 oxygen atoms in 6 CO molecules and 3 H O molecules, we need 15/2 Omolecules in order to have 15 oxygen atoms on the reactant side.

6 6 2 2 2C H (l) + O (g) ÷ 6 CO (g) + 3 H O(g); Multiply by two to give whole numbers.

6 6 2 2 22 C H (l) + 15 O (g) ÷ 12 CO (g) + 6 H O(g)

4 10 2 2 2b. The formulas of the reactants and products are C H (g) + O (g) ÷ CO (g) + H O(g).

4 10 2 2 2C H (g) + O (g) ÷ 4 CO (g) + 5 H O(g); Multiply by two to give whole numbers.

4 10 2 2 22 C H (g) + 13 O (g) ÷ 8 CO (g) + 10 H O(g)

12 22 11 2 2 2c. C H O (s) + 12 O (g) ÷ 12 CO (g) + 11 H O(g)

2 2 3 2 2 3d. 2 Fe(s) + O (g) ÷ Fe O (s); For whole numbers: 4 Fe(s) + 3 O (g) ÷ 2 Fe O (s)

2 2 3e. 2 FeO(s) + O (g) ÷ Fe O (s); For whole numbers, multiply by two.

2 2 34 FeO(s) + O (g) ÷ 2 Fe O (s)

8 2 380. a. 16 Cr(s) + 3 S (s) ÷ 8 Cr S (s)


3 2 3 2 2b. 2 NaHCO (s) ÷ Na CO (s) + CO (g) + H O(g)

3 2c. 2 KClO (s) ÷ 2 KCl(s) + 3 O (g)

3 2d. 2 Eu(s) + 6 HF(g) ÷ 2 EuF (s) + 3 H (g)

281. a. SiO (s) + C(s) ÷ Si(s) + CO(g)

2Balance oxygen atoms: SiO + C ÷ Si + 2 CO

2Balance carbon atoms: SiO (s) + 2 C(s) ÷ Si(s) + 2 CO(g)

4 2b. SiCl (l) + Mg(s) ÷ Si(s) + MgCl (s)

4 2Balance Cl atoms: SiCl + Mg ÷ Si + 2 MgCl

4 2Balance Mg atoms: SiCl (l) + 2 Mg(s) ÷ Si(s) + 2 MgCl (s)

2 6c. Na SiF (s) + Na(s) ÷ Si(s) + NaF(s)

2 6Balance F atoms: Na SiF + Na ÷ Si + 6 NaF

2 6Balance Na atoms: Na SiF (s) + 4 Na(s) ÷ Si(s) + 6 NaF(s)

82. Unbalanced equation:

2 3 4 2 2 4 3 4 4 2CaF C3Ca (PO ) (s) + H SO (aq) ÷ H PO (aq) + HF(aq) + CaSO C2H O(s)

4Balancing Ca , F , and PO :2+ - 3-

2 3 4 2 2 4 3 4 4 2CaF C3Ca (PO ) (s) + H SO (aq) ÷ 6 H PO (aq) + 2 HF(aq) + 10 CaSO C2H O(s)

On the right-hand side there are 20 extra hydrogen atoms, 10 extra sulfates, and 20 extra watermolecules. We can balance the hydrogen and sulfate with 10 sulfuric acid molecules. The extrawaters came from the water in the sulfuric acid solution. The balanced equation is:

2 3 4 2 2 4 2 3 4 4 2CaF C3Ca (PO ) (s) + 10 H SO (aq) + 20 H O(l) ÷ 6 H PO (aq) + 2 HF(aq) + 10 CaSO C2H O(s)

12 22 11 2 2 5 283. C H O (aq) + H O(l) ÷ 4 C H OH(aq) + 4 CO (g)

3 2 4 284. CaSiO (s) + 6 HF(aq) ÷ CaF (aq) + SiF (g) + 3 H O(l)

Reaction Stoichiometry

85. The stepwise method to solve stoichiometry problems is outlined in the text. Instead of calculatingintermediate answers for each step, we will combine conversion factors into one calculation. This


practice reduces round-off error and saves time.

4 2 2 7 2 3 2 2The balanced reaction is: (NH ) Cr O (s) ÷ Cr O (s) + N (g) + 4 H O(g)

4 2 2 7 4 2 2 710.8 g (NH ) Cr O × = 4.28 × 10 mol (NH ) Cr O-2

4 2 2 7 2 34.28 × 10 mol (NH ) Cr O × = 6.51 g Cr O-2

4 2 2 7 24.28 × 10 mol (NH ) Cr O × = 1.20 g N-2

4 2 2 7 24.28 × 10 mol (NH ) Cr O × = 3.09 g H O-2

2 3 2 386. Fe O (s) + 2 Al(s) ÷ 2 Fe(l) + Al O (s)

15.0 g Fe × = 0.269 mol Fe; 0.269 mol Fe × = 7.26 g Al

2 30.269 mol Fe × = 21.5 g Fe O

2 30.269 mol Fe × = 13.7 g Al O

87. 1.000 kg Al × = 4355 g

2 2 4 2 2 388. a. Ba(OH) C8H O(s) + 2 NH SCN(s) ÷ Ba(SCN) (s) + 10 H O(l) + 2 NH (g)

2 2b. 6.5 g Ba(OH) C8H O × = 0.0206 mol = 0.021 mol

2 2 40.021 mol Ba(OH) C8H O × = 3.2 g NH SCN

89. 1.0 ton CuO ×

= 7.2 × 10 g coke4

90. 1.0 × 10 kg waste × 4


4 = 3.4 × 10 g tissue if all NH converted4 +

4Since only 95% of the NH ions react:+

mass of tissue = (0.95) (3.4 × 10 g) = 3.2 × 10 g or 32 kg bacterial tissue4 4

91. a. Molar mass = 195.1 + 2(14.01) + 6(1.008) + 2(35.45) = 300.1 g/mol

% Pt = × 100 = 65.01% Pt; % N = × 100 = 9.337% N

% H = × 100 = 2.015% H; % Cl = × 100 = 23.63% Cl

65.01% Pt; 9.337% N; 2.015% H; 23.63% Cl

2 4 b. 100. g K PtCl ×

3 2 2 = 72.3 g Pt(NH ) Cl

2 4100. g K PtCl × = 35.9 g KCl

7 6 3 4 6 392. a. 1.00 × 10 g C H O × = 73.9 g C H O2

7 6 3b. 1.00 × 10 g C H O × 2

= 1.30 × 10 g aspirin2

Limiting Reactants and Percent Yield

2 293. a. Mg(s) + I (s) ÷ MgI (s)

2From the balanced equation, 100 molecules of I reacts completely with 100 atoms of Mg. Wehave a stoichiometric mixture. Neither is limiting.

2b. 150 atoms Mg × = 150 molecules I needed

2We need 150 molecules I to react completely with 150 atoms Mg; we only have 100 molecules

2 2I . Therefore, I is limiting.

2. 2c. 200 atoms Mg × = 200 molecules I ; Mg is limiting since 300 molecules I are present.

2 2d. 0.16 mol Mg × = 0.16 mol I ; Mg is limiting since 0.25 mol I are present.


2e. 0.14 mol Mg × = 0.14 mol I needed; Stoichiometric mixture. Neither is limiting.

2 2 2f. 0.12 mol Mg × = 0.12 mol I needed; I is limiting since only 0.08 mol I are present.

2g. 6.078 g Mg × = 63.46 g I

Stoichiometric mixture. Neither is limiting.

2h. 1.00 g Mg × = 10.4 g I

2 210.4 g I needed, but we only have 2.00 g. I is limiting.

2i. From h above, we calculated that 10.4 g I will react completely with 1.00 g Mg. We have 20.00

2 2g I . I is in excess. Mg is limiting.

2 2 294. 2 H (g) + O (g) ÷ 2 H O(g)

2 2a. 50 molecules H × = 25 molecules O

Stoichiometric mixture. Neither is limiting.

2 2 2b. 100 molecules H × = 50 molecules O ; O is limiting since only 40 molecules

2 O are present.

2 2c. From b, 50 molecules of O will react completely with 100 molecules of H . We have 100

2 2molecules (an excess) of O . So, H is limiting.

2 2 2 2d. 0.50 mol H × = 0.25 mol O ; H is limiting since 0.75 mol O are present.

2 2 2 2e. 0.80 mol H × = 0.40 mol O ; H is limiting since 0.75 mol O are present.

2 2f. 1.0 g H × = 0.25 mol O

Stoichiometric mixture, neither is limiting.

2 2 2g. 5.00 g H × = 39.7 g O ; H is limiting since

2 56.00 g O are present.

95. a. 10.0 g Hg × = 4.99 × 10 mol Hg-2


2 29.00 g Br × = 5.63 × 10 mol Br-2

2The required mol ratio from the balanced equation is 1 mol Br to 1 mol Hg. The actual mol ratiois:

= 1.13

This is higher than the required ratio, so Hg is the limiting reagent.

24.99 × 10 mol Hg × = 18.0 g HgBr produced-2

24.99 × 10 mol Hg × = 7.97 g Br reacted-2

2 2 2 2excess Br = 9.00 g Br - 7.97 g Br = 1.03 g Br

b. 5.00 mL Hg × = 0.339 mol Hg

2 25.00 mL Br × = 0.0970 mol Br

2 2Br is limiting since the actual moles of Br present is well below the required 1:1 mol ratio.

2 20.0970 mol Br × = 35.0 g HgBr produced

2 296. 1.50 g BaO × = 8.86 × 10 mol BaO-3

25.0 mL × = 1.87 × 10 mol HCl-2

2The required mol ratio from the balanced reaction is 2 mol HCl to 1 mol BaO . The actual ratio is:

= 2.11

2Since the actual mol ratio is larger than the required mol ratio, the denominator (BaO ) is the limitingreagent.

2 2 28.86 × 10 mol BaO × = 0.301 g H O-3

The amount of HCl reacted is:


28.86 × 10 mol BaO × = 1.77 × 10 mol HCl-3 -2

excess mol HCl = 1.87 × 10 mol - 1.77 × 10 mol = 1.0 × 10 mol HCl-2 -2 -3

mass of excess HCl = 1.0 × 10 mol HCl × = 3.6 × 10 g HCl-3 -2

3 4 2 2 4 4 3 497. Ca (PO ) + 3 H SO ÷ 3 CaSO + 2 H PO

3 4 2 3 4 21.0 × 10 g Ca (PO ) × = 3.2 mol Ca (PO )3

2 4 2 41.0 × 10 g conc. H SO × = 10. mol H SO3

2 4 3 4 2The required mol ratio from the balanced equation is 3 mol H SO to 1 mol Ca (PO ) . The actual

ratio is: = 3.1

3 4 2This is higher than the required mol ratio, so Ca (PO ) is the limiting reagent.

3 4 2 43.2 mol Ca (PO ) × = 1300 g CaSO produced

3 4 2 3 43.2 mol Ca (PO ) × = 630 g H PO produced

98. An alternative method to solve limiting reagent problems is to assume each reactant is limiting andcalculate how much product could be produced from each reactant. The reactant that produces thesmallest amount of product will run out first and is the limiting reagent.

35.00 × 10 g NH × = 2.94 × 10 mol HCN6 5

25.00 × 10 g O × = 1.04 × 10 mol HCN6 5

45.00 × 10 g CH × = 3.12 × 10 mol HCN6 5

2O is limiting since it produces the smallest amount of HCN. Although more product could be

3 4 2produced from NH and CH , only enough O is present to produce 1.04 × 10 mol HCN. The5

mass of HCN produced is:

1.04 × 10 mol HCN × = 2.81 × 10 g HCN5 6

2 25.00 × 10 g O × = 5.63 × 10 g H O6 6


2 6 2 2 599. C H (g) + Cl (g) ÷ C H Cl(g) + HCl(g)

2 6 2 6 2 2300. g C H × = 9.98 mol C H ; 650. g Cl × = 9.17 mol Cl

2 6The balanced equation requires a 1:1 mol ratio between reactants. 9.17 mol of C H will react with

2 2 6 2all of the Cl present (9.17 mol). Since 9.98 mol C H is present, Cl is the limiting reagent.

2 5The theoretical yield of C H Cl is:

2 2 59.17 mol Cl × = 592 g C H Cl

Percent yield = × 100 = 82.8%

7 6 3 4 6 3 9 8 4 2 3 2100. C H O + C H O ÷ C H O + HC H O

7 6 3 7 6 31.50 g C H O × = 1.09 × 10 mol C H O-2

4 6 3 4 6 32.00 g C H O × = 1.96 × 10 mol C H O-2

7 6 3 7 6 3C H O is the limiting reagent since the actual moles of C H O are below the required 1:1 mol ratio.The theoretical yield of aspirin is:

7 6 3 9 8 41.09 × 10 mol C H O × = 1.96 g C H O-2

% yield = × 100 = 76.5%

3 3101. 2.50 metric tons Cu FeS ×

= 1.39 × 10 g Cu (theoretical)6

1.39 × 10 g Cu (theoretical) × = 1.20 × 10 g Cu = 1.20 × 10 kg Cu6 6 3

= 1.20 metric tons Cu (actual)

4 2 3 3102. P (s) + 6 F (g) ÷ 4 PF (g); The theoretical yield of PF is:

3 3 120. g PF (actual) × = 154 g PF (theoretical)


3 2154 g PF × × × = 99.8 g F

2 399.8 g F are needed to produce an actual PF yield of 78.1%.

Additional Exercises

103.

The atomic mass is 54.94 amu. From the periodic table, the element is manganese (Mn).

2104. In one hour, the 1000. kg of wet cereal produced contains 580 kg H O and 420 kg of cereal. We want

2 2the final product to contain 20.% H O. Let x = mass of H O in final product.

2 = 0.20, x = 84 + 0.20 x, x = 105 . 110 kg H O

The amount of water to be removed is 580 - 110 = 470 kg/hr.

105. Empirical formula mass = 12.01 + 1.008 = 13.02 g/mol; Since 104.14/13.02 = 7.998 . 8, the

8 8 8molecular formula for styrene is (CH) = C H .

8 82.00 g C H × × × = 9.25 × 10 atoms H22

2106. 41.98 mg CO × = 11.46 mg C; %C = × 100 = 57.85% C

26.45 mg H O × = 0.722 mg H; %H = × 100 = 3.64% H

%O = 100.00 - (57.85 + 3.64) = 38.51% O

Out of 100.00 g terephthalic acid, there are:

57.85 g C × = 4.817 mol C; 3.64 g H × = 3.61 mol H

38.51 g O × = 2.407 mol O

= 2.001; = 1.50; = 1.000

4 3 2C:H:O mol ratio is 2:1.5:1 or 4:3:2. Empirical formula: C H O


4 3 2Mass of C H O . 4(12) + 3(1) + 2(16) = 83

8 6 4Molar mass = = 166 g/mol; = 2; Molecular formula: C H O

107. 17.3 g H × = 17.2 mol H; 82.7 g C × = 6.89 mol C

2 5 = 2.50; The empirical formula is C H .

The empirical formula mass is ~29 g, so two times the empirical formula would put the compound in

2 5 2 4 10the correct range of the molar mass. Molecular formula = (C H ) = C H

4 102.59 × 10 atoms H × = 4.30 × 10 mol C H23 -2

4 10 4 104.30 × 10 mol C H × = 2.50 g C H-2

3 8108. Assuming 100.00 g E H :

mol E = 8.73 g H × = 3.25 mol E

, x = molar mass of E = 28.1 g/mol; atomic mass of E = 28.1 amu

2 4 2 4 2109. Mass of H O = 0.755 g CuSO CxH O - 0.483 g CuSO = 0.272 g H O

4 40.483 g CuSO × = 0.00303 mol CuSO

2 20.272 g H O × = 0.0151 mol H O

4 2; Compound formula = CuSO C5H O, x = 5

110. a. Only acrylonitrile contains nitrogen. If we have 100.00 g of polymer:

3 38.80 g N × = 33.3 g C H N

3 3 3 3% C H N = = 33.3% C H N


2Only butadiene in the polymer reacts with Br :

2 4 60.605 g Br × = 0.205 g C H

4 6 4 6% C H = × 100 = 17.1% C H

b. If we have 100.0 g of polymer:

3 3 3 333.3 g C H N × = 0.628 mol C H N

4 6 4 617.1 g C H × = 0.316 mol C H

8 8 8 849.6 g C H × = 0.476 mol C H

Dividing by 0.316:

This is close to a mol ratio of 4:2:3. Thus, there are 4 acrylonitrile to 2 butadiene to 3 styrene

4 2 3molecules in the polymer or (A B S ).

2 24 30 3111. 1.20 g CO × = 0.428 g C H N O

24 30 3 × 100 = 42.8% C H N O

4 2 2 4 8 2 2112. a. CH (g) + 4 S(s) ÷ CS (l) + 2 H S(g) or 2 CH (g) + S (s) ÷ 2 CS (l) + 4 H S(g)

4 4b. 120. g CH × = 7.48 mol CH ; 120. g S × = 3.74 mol S

4 4The required S to CH mol ratio is 4:1. The actual S to CH mol ratio is:

= 0.500

This is well below the required ratio so sulfur is the limiting reagent.

2 2The theoretical yield of CS is: 3.74 mol S × = 71.2 g CS

2 8The same amount of CS would be produced using the balanced equation with S .


113. 453 g Fe ×

2 3mass % Fe O =

2114. a. Mass of Zn in alloy = 0.0985 g ZnCl × = 0.0473 g Zn

%Zn = × 100 = 9.34% Zn; %Cu = 100.00 - 9.34 = 90.66% Cu

b. The Cu remains unreacted. After filtering, washing, and drying, the mass of the unreacted coppercould be measured.

115. Assuming one mol of vitamin A (286.4 g Vitamin A):

mol C = 286.4 g Vitamin A × = 20.00 mol C

mol H = 286.4 g Vitamin A × = 30.00 mol H

Since one mol of Vitamin A contains 20 mol C and 30 mol H, the molecular formula of Vitamin A

20 30is C H E. To determine E, let’s calculate the molar mass of E.

286.4 g = 20(12.01) + 30(1.008) + molar mass E, molar mass E = 16.0 g/mol

20 30From the periodic table, E = oxygen and the molecular formula of Vitamin A is C H O.

Challenge Problems

116. ; Assuming 100 atoms, let x = number of Rb atoms and 100 - x = number of Rb85 87

atoms.

= 2.591, x = 259.1 - 2.591 x, = 72.15% Rb85

0.7215 (84.9117) + 0.2785 (A) = 85.4678, A = = 86.92 amu = atomic mass of Rb 87

2117. First, we will determine composition in mass percent. We assume all the carbon in the 0.213 g CO

2came from 0.157 g of the compound and that all the hydrogen in the 0.0310 g H O came from the 0.157 g of the compound.


20.213 g CO × = 0.0581 g C; %C = × 100 = 37.0% C

20.0310 g H O × = 3.47 × 10 g H; %H = = 2.21% H-3

We get %N from the second experiment:

30.0230 g NH × = 1.89 × 10 g N-2

%N = × 100 = 18.3% N

The mass percent of oxygen is obtained by difference:

%O = 100.00 - (37.0 + 2.21 + 18.3) = 42.5%

So out of 100.00 g of compound, there are:

37.0 g C × = 3.08 mol C; 2.21 g H × = 2.19 mol H

18.3 g N × = 1.31 mol N; 42.5 g O × = 2.66 mol O

The last, and often the hardest part, is to find simple whole number ratios. Divide all mole values bythe smallest number:

= 2.35; = 1.67; = 1.00; = 2.03

7 5 3 6Multiplying all these ratios by 3 gives an empirical formula of C H N O .

3 3118. 1.0 × 10 kg HNO × = 1.6 × 10 mol HNO6 7

3 3We need to get the relationship between moles of HNO and moles of NH . We have to use all 3equations.

3 3Thus, we can produce 16 mol HNO for every 24 mol NH we begin with:

31.6 × 10 mol HNO × = 4.1 × 10 g or 4.1 × 10 kg7 8 5

This is an oversimplified answer. In practice, the NO produced in the third step is recycled back


continuously into the process in the second step. If this is taken into consideration, then the conversion

3 3 3factor between mol NH and mol HNO turns out to be 1:1, i.e., 1 mol of NH produces 1 mol of

3HNO . Taking into consideration that NO is recycled back gives an answer of

32.7 × 10 kg NH reacted.5

119. Total mass of copper used:

10,000 boards × = 6.9 × 10 g Cu5

Amount of Cu removed = 0.80 × 6.9 × 10 g = 5.5 × 10 g Cu5 5

5.5 × 10 g Cu × 5

3 4 2= 1.8 × 10 g Cu(NH ) Cl6

35.5 × 10 g Cu × = 5.9 × 10 g NH5 5

120. a. From the reaction stoichiometry we would expect to produce 4 mol of acetaminophen for every

6 5 34 mol of C H O N reacted. The actual yield is 3 moles of acetaminophen compared to atheoretical yield of 4 moles of acetaminophen. Solving for percent yield by mass (where M =molar mass acetaminophen):

% yield = × 100 = 75%

b. The product of the percent yields of the individual steps must equal the overall yield, 75%.

(0.87) (0.98) (x) = 0.75, x = 0.88; Step III has a % yield = 88%.

2 2 4 2 4 4121. 10.00 g XCl + excess Cl ÷ 12.55 g XCl ; 2.55 g Cl reacted with XCl to form XCl . XCl contains

2 22.55 g Cl and 10.00 g XCl . From mol ratios, 10.00 g XCl must also contain 2.55 g Cl; mass X in

2 XCl = 10.00 - 2.55 = 7.45 g X.

2.55 g Cl × = 3.60 × 10 mol X-2

So, 3.60 × 10 mol X must equal 7.45 g X. The molar mass of X is:-2

; Atomic mass = 207 amu so X is Pb.

2 3 2122. 4.000 g M S ÷ 3.723 g MO

2 2 3There must be twice as many mol of MO as mol of M S in order to balance M in the reaction.

2 2 3Setting up an equation for 2 mol MO = mol M S where A = molar mass M:


2 ,

8.000 A + 256.0 = 7.446 A + 358.2, 0.554 A = 102.2, A = 184 g/mol; atomic mass = 184 amu

123. Consider the case of aluminum plus oxygen. Aluminum forms Al ions; oxygen forms O anions.3+ 2-

2 3The simplest compound of the two elements is Al O . Similarly, we would expect the formula of any

2 3group 6A element with Al to be Al X . Assuming this, out of 100.00 g of compound there are 18.56g Al and 81.44 g of the unknown element, X. Let’s use this information to determine the molar massof X which will allow us to identify X from the periodic table.

18.56 g Al × = 1.032 mol X

81.44 g of X must contain 1.032 mol of X.

The molar mass of X = = 78.91 g/mol X.

2 3From the periodic table, the unknown element is selenium and the formula is Al Se .

124. NaCl(aq) + Ag (aq) ÷ AgCl(s); KCl(aq) + Ag (aq) ÷ AgCl(s)+ +

8.5904 g AgCl × = 5.991 × 10 mol Cl-2 -

The molar masses of NaCl and KCl are 58.44 and 74.55 g/mol, respectively. Let x = g NaCland y = g KCl:

x + y = 4.000 g and = 5.991 × 10 total mol Cl or 1.276 x + y = 4.466-2 -

Solving using simultaneous equations:

1.276 x + y = 4.466 - x - y = -4.0000.276 x = 0.466, x = 1.69 g NaCl and y = 2.31 g KCl

% NaCl = × 100 = 42.3% NaCl; % KCl = 57.7%

125. The balanced equations are:

3 2 2 3 2 2 24 NH (g) + 5 O (g) ÷ 4 NO(g) + 6 H O(g) and 4 NH (g) + 7 O (g) ÷ 4 NO (g) + 6 H O(g)

2Let 4x = number of mol of NO formed, and let 4y = number of mol of NO formed. Then:


3 2 2 3 2 2 24x NH + 5x O ÷ 4x NO + 6x H O and 4y NH + 7y O ÷ 4y NO + 6y H O

3 2All the NH reacted, so 4x + 4y = 2.00. 10.00 - 6.75 = 3.25 mol O reacted, so 5x + 7y = 3.25.

Solving by the method of simultaneous equations:

20 x + 28 y = 13.0 -20 x - 20 y = -10.0

8 y = 3.0, y = 0.38; 4x + 4 × 0.38 = 2.00, x = 0.12

mol NO = 4x = 4 × 0.12 = 0.48 mol NO formed

CHAPTER THREE STOICHIOMETRY - … · 29 CHAPTER THREE STOICHIOMETRY Questions 18. The two major isotopes of boron are 10B and 11B.The listed mass of 10.81 is the average mass of a

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