chapter 9 calculations from chemical …faculty.chemeketa.edu/lemme/ch...

CHAPTER 9

CALCULATIONS FROM CHEMICAL EQUATIONSSOLUTIONS TO REVIEW QUESTIONS

1. A mole ratio is the ratio between the mole amounts of two atoms and/or moleculesinvolved in a chemical reaction.

2. In order to convert grams to moles the molar mass of the compound under considerationneeds to be determined.

3. The balanced equation is

(a) Correct:

(b) Incorrect: would produce

(c) Correct: see equation

(d) Correct: see equation

(e) Incorrect: requires to produce

(f) Correct: will react with ( are present in excess)and will be formed.

(g) Incorrect:

The amount of water present (100. g) is less than needed to react withis the limiting reactant.

(h) Incorrect: water is the limiting reactant.

1100. g H2O2a 1 mol

18.02 gb ¢ 2 mol PH3

6 mol H2O≤ a33.99 g

molb = 62.9 g PH3

H2O200. g Ca3P2.

(200. g Ca3P2)a 1 mol

182.2 gb ¢ 6 mol H2O

1 mol Ca3P2≤ a 18.02 g

molb = 119 g H2O

12 mol Ca3P22¢3 mol Ca(OH)2

1 mol Ca3P2≤ = 6 mol Ca(OH)2

6 mol Ca(OH)2

3 mol H2O12 mol H2O2 mol Ca3P2

12 mol Ca3P22¢ 6 mol H2O

1 mol Ca3P2≤ = 12 mol H2O

4.0 mol PH3.12 mol H2O2 mol Ca3P2

(1 g Ca3P2)a 1 mol

182.2 gb ¢ 2 mol PH3

1 mol Ca3P2≤ a33.99 g

molb = 0.4 g PH3

0.4 g PH31 g Ca3P2

11 mol Ca3P22¢ 2 mol PH3

1 mol Ca3P2≤ = 2 mol PH3

Ca3P2 + 6 H2O ¡ 3 Ca1OH22 + 2 PH3

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(a) Correct

(b) Incorrect: (not 12 mol HCN)

(c) Correct

(d) Incorrect: (not )

(e) Correct

(f) Incorrect: is the limiting reactant

(not 3 mol HCN)

5. The theoretical yield of a chemical reaction is the maximum amount of product that canbe produced based on a balanced equation. The actual yield of a reaction is the actualamount of product obtained.

6. You can calculate the percent yield of a chemical reaction by dividing the actual yield bythe theoretical yield and multiplying by one hundred.

13 mol O22¢2 mol HCN

3 mol O2≤ = 2 mol HCN

O2

4 mol H2O112 mol HCN2¢ 6 mol H2O

2 mol HCN≤ = 36 mol H2O

116 mol O22¢ 2 mol HCN

3 mol O2≤ = 10.7 mol HCN

2 CH4 + 3 O2 + 2 NH3 ¡ 2 HCN + 6 H2O

- 96 -

- Chapter 9 -


CHAPTER 9

SOLUTIONS TO EXERCISES

1. (a)

(b)

(c)

(d) The conversion is:

2. (a)

(b)

(c)

(d)

3. (a)

(b)

(c)

(d)

(e) 1500.0 mL Br22a3.119 g

mLb = 1559.5 g Br2 = 1.560 * 103 g Br2

172 mmol HCl2a 1 mol

1000 mmolb a36.46 g

molb = 2.6 g HCl

110.5 mol NH32a17.03 g

molb = 179 g NH3

1125 kg CaCO32a1000 g

kgb = 1.25 * 105 g CaCO3

12.55 mol Fe(OH)32a106.9 g

molb = 273 g Fe(OH)3

1250 mL C2H5OH2a0.789 g

mLb a 1 mol

46.07 gb = 4.3 mol C2H5OH

19.8 * 1024 molecules CO22¢ 1 mol

6.022 * 1023 molecules≤ = 16 mol CO2

1525 mg ZnCl22a 1 g

1000 mgb a 1 mol

136.3 gb = 3.85 * 10-3 mol ZnCl2

12.10 kg NaHCO32a1000 g

kgb a 1 mol

84.01 gb = 25.0 mol NaHCO3

116.8 mL solution2a1.727 g

mLb ¢0.800 g H2SO4

g solution≤ a 1 mol

98.09 gb = 0.237 mol H2SO4

mL sol ¡ g sol ¡ g H2SO4 ¡ mol H2SO4

15.4 * 102 g (NH4)2C2O42a 1 mol

124.1 gb = 4.4 mol (NH4)2C2O4

156 mmol NaOH2a 1 mol

1000 mmolb = 0.056 mol NaOH

125.0 g KNO32a 1 mol

101.1 gb = 0.247 mol KNO3

- 97 -


4. (a)

(b)

(c)

(d)

(e)

5. Larger number of molecules:

Water has a lower molar mass than hydrogen peroxide. 10.0 grams of water has a lowermolar mass, contains more moles, and therefore more molecules than 10.0 g of

6. Larger number of molecules: 25.0 g HCl or

HCl contains more molecules

7. Mole ratios

(a) (d)

(b) (e)

(c) (f)8 mol H2O

9 mol O2

9 mol O2

6 mol CO2

6 mol CO2

8 mol H2O

2 mol C3H7OH

9 mol O2

8 mol H2O

2 mol C3H7OH

6 mol CO2

2 mol C3H7OH

2 C3H7OH + 9 O2 ¡ 6 CO2 + 8 H2O

= 2.84 * 1023 molecules C6H12O6

185.0 g C6H12O62a 1 mol

180.2 gb a6.022 * 1023 molecules

molb

125.0 g HCl2a 1 mol

36.46 gb a6.022 * 1023 molecules

molb = 4.13 * 1023 molecules HCl

85 g C6H12O6

H2O2.

10.0 g H2O or 10.0 g H2O2

(75 mL solution2a1.175 g

mLb a 0.200 g K2CrO4

g solutionb = 18 g K2CrO4

= 1.35g C6H12O6

(4.50 * 1021 molecules C6H12O6)¢ 1 mol

6.022 * 1023 molecules≤ a 180.2 g

molb

(0.725 mol Bi2S3)a514.2 g

molb = 373 g Bi2S3

(0.0600 mol HC2H3O2)a 60.05 g

molb = 3.60 g HC2H3O2

(0.00844 mol NiSO4)a 154.8 g

molb = 1.31 g NiSO4

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- Chapter 9 -


8. Mole ratios

(a) (d)

(b) (e)

(c) (f)

9.



(a)

(b)

12.

(a)

(b) 10.600 mol CH42¢4 mol Al(OH)3

3 mol CH4≤ = 0.800 mol Al(OH)3

1100. g Al4C32a 1 mol

144.0 gb ¢ 12 mol H2O

1 mol Al4C3≤ = 8.33 mol H2O

Al4C3 + 12 H2O ¡ 4 Al(OH)3 + 3 CH4

11.25 mol H2O2¢1 mol MnCl22 mol H2O

≤ = 0.625 mol MnCl2s

11.05 mol MnO22¢ 4 mol HCl

1 mol MnO2≤ = 4.20 mol HCl

MnO2(s) + 4 HCl(aq) ¡ Cl2(g) + MnCl2(aq) + 2 H2O(l)

15.60 mol HCl2¢ 2 mol Cl24 mol HCl

≤ = 2.80 mol Cl2

4 HCl + O2 ¡ 2 Cl2 + 2 H2O

17.75 mol C2H5OH2¢ 2 mol CO2

1 mol C2H5OH≤ = 15.5 mol CO2

C2H5OH + 3 O2 ¡ 2 CO2 + 3 H2O

2 mol H3PO4

6 mol HCl

3 mol CaCl2

2 mol H3PO4

6 mol HCl

1 mol Ca3(PO4)2

6 mol HCl

2 mol H3PO4

1 mol Ca3(PO4)2

2 mol H3PO4

3 mol CaCl2

1 mol Ca3(PO4)2

3 CaCl2 + 2 H3PO4 ¡ Ca3(PO4)2 + 6 HCl

- Chapter 9 -

- 99 -


13. Grams of NaOH

The conversion is:

14. Grams of

The conversion is:


The conversion is:


Calculate the grams of both and Fe to produce


(a)

(b)

(c)

(d) 12.75 mol of H2O2¢ 4 mol CO2

6 mol H2O≤ a 44.01 g

molb = 80.7 g CO2

175.0 g C2H62a 1 mol

30.07 gb ¢ 4 mol CO2

2 mol C2H6≤ a 44.01 g

molb = 2.20 * 102 g CO2

18.00 g H2O2a 1 mol

18.02 gb ¢ 4 mol CO2

6 mol H2O≤ a44.01 g

molb = 13.0 g CO2

115.0 mol C2H62¢ 7 mol O2

2 mol C2H6≤ = 52.5 mol O2

2 C2H6 + 7 O2 ¡ 4 CO2 + 6 H2O

1375 g Fe3O42a 1 mol

231.6 gb ¢ 3 mol Fe

1 mol Fe3O4≤ a 55.85 g

molb = 271 g Fe

1375 g Fe3O42¢ 1 mol

231.6 g≤ a 4 mol H2O

1 mol Fe3O4≤ a 18.02 g

molb = 117 g H2O

375 g Fe3O4H2O

3 Fe + 4 H2O ¡ Fe3O4 + 4 H2

1125 kg Fe2O32a 1 kmol

159.7 kgb a 2 kmol Fe

1 kmol Fe2O3b a 55.85 kg

kmolb = 87.4 kg Fe

kg Fe2O3 ¡ kmol Fe2O3 ¡ kmol Fe ¡ kg Fe

Fe2O3 + 3 C ¡ 2 Fe + 3 CO

110.0 g Zn2a 1 mol

65.39 gb ¢1 mol Zn3(PO4)2

3 mol Zn≤ a386.1 g

molb = 19.7 g Zn3(PO4)2

g Zn ¡ mol Zn ¡ mol Zn3(PO4)2 ¡ g Zn3(PO4)2

3 Zn + 2 H3PO4 ¡ Zn3(PO4)2 + 3 H2

Zn3(PO4)2

1500 g Ca(OH)22a 1 mol

74.10 gb a 2 mol NaOH

1 mol Ca(OH)2b a 40.00 g

molb = 5 * 102 g NaOH

g Ca(OH)2 ¡ mol Ca(OH)2 ¡ mol NaOH ¡ g NaOH

Ca(OH)2 + Na2CO3 ¡ 2 NaOH + CaCO3

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- Chapter 9 -


(e)

(f)

18.

(a)

(b)

(c)

(d)

(e)

(f)

19. Hydrogen(a) Oxygen

Hydrogen is the limiting reactant.

1221 g Fe2O32a 1 mol

159.7 gb ¢ 4 mol FeS2

2 mol Fe2O3≤ a 120.0 g

molb = 332 g FeS2

140.6 g SO22a 1 mol

64.07 gb ¢ 11 mol O2

8 mol SO2≤ = 0.871 mol O2

10.512 mol FeS22¢ 8 mol SO2

4 mol FeS2≤ a64.07 g

molb = 65.6 g SO2

11.55 mol Fe2O32¢ 8 mol SO2

2 mol Fe2O3≤ = 6.20 mol SO2

14.50 mol FeS22¢ 11 mol O2

4 mol FeS2≤ = 12.4 mol O2

11.00 mol FeS22a2 mol Fe2O3

4 mol FeS2b = 0.500 mol Fe2O3

4 FeS2 + 11 O2 ¡ 2 Fe2O3 + 8 SO2

1125 g H2O2¢ 1 mol

18.02 g≤ ¢2 mol C2H6

6 mol H2O≤ a30.07 g

molb = 69.5 g C2H6

125.0 mol C2H62¢ 7 mol O2

2 mol C2H6≤ a 32.00 g

mol O2b = 2.80 * 103 g O2

- Chapter 9 -

- 101 -


Potassium(a) Chlorine

Potassium is the limiting reactant.

Silver(b) Chlorine

Silver is the limiting reactant.

Lithium(a) Iodine

No limiting reactant.

Hydrogen(b) Bromine

Bromine is the limiting reactant.

- 102 -

- Chapter 9 -

20.

21.


23. (a)

16.0 g 12.0 gChoose one of the products and calculate its mass that would be produced fromeach given reactant. Using as the product:

Since produces less it is the limiting reactant and KOH is in excess.KNO3,HNO3

112.0 g HNO32a 1 mol

63.02 gb ¢1 mol KNO3

1 mol KOH≤ a 101.1 g

molb = 19.3 g KNO3

116.0 g KOH2¢ 1 mol

56.10 g≤ a1 mol KNO3

1 mol KOH≤ a 101.1 g

molb = 28.8 g KNO3

KNO3

H2O+KNO3¡HNO3+KOH

HydrogenIron(b) Oxygen

Water is the limiting reactant.

Nitrogen(a) Oxygen

Oxygen is the limiting reactant.

Aluminum(b) Oxygen

Oxygen is the limiting reactant.

- Chapter 9 -

- 103 -

22.


(b)

10.0 g 10.0 g

Choose one of the products and calculate its mass that would be produced fromeach given reactant. Using as the product:

Since produces less it is the limiting reactant and NaOH is in excess.

24. (a) 3

50.0 g 6.00 g


Since produces less it is the limiting reactant and is in excess.

(b)

40.0 g 16.0 g


Since produces less it is the limiting reactant and Fe is in excess.

25. Limiting reactant calculations

(a) Reaction between and Convert each amount to moles of CO2

20.0 g O220.0 g C3H8

C3H8 + 5 O2 ¡ 3 CO2 + 4 H2O

H2,H2O

116.0 g H2O2a 1 mol

18.02 gb ¢ 4 mol H2

4 mol H2O≤ a2.016 g

molb = 1.79 g H2

(40.0 g Fe)a 1 mol

55.85 gb ¢4 mol H2

3 mol Fe≤ a2.016 g

molb = 1.93 g H2

H2

4 H2+Fe3O4¡4 H2O+3 Fe

Bi(NO3)3Bi2S3,H2S

(6.00 g H2S)a 1 mol

34.09 gb ¢1 mol Bi2S3

3 mol H2S≤ a514.2 g

molb = 30.2 g Bi2S3

150.0 g Bi(NO3)3)a 1 mol

395.0 gb a 1 mol Bi2S3

2 mol Bi(NO3)3b a 514.2 g

molb = 32.5 g Bi2S3

Bi2S3

6 HNO3+Bi2S3¡H2S+2 Bi(NO3)3

H2O,H2SO4

110.0 g H2SO42a 1 mol

98.09 gb ¢ 2 mol H2O

1 mol H2SO4≤ a 18.02 g

molb = 3.67 g H2O

110.0 g NaOH2a 1 mol

40.00 gb ¢ 2 mol H2O

2 mol NaOH≤ a18.02 g

molb = 4.51 g H2O

H2O

2 H2O+Na2SO4¡H2SO4+2 NaOH

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HEINS09-095-117v4.qxd 12/30/06 1:58 PM 4

is the limiting reactant. The yield is

(b) Reaction between and Convert each amount to moles of

is the limiting reactant. The yield is

(c) Reaction between and According to the equation, will react with Therefore,is the limiting reactant and will remain unreacted.

When the reaction is completed, and will be inthe container.

26. The balanced equation is 2 (a) Reaction between and .

Convert each amount to moles of

The is the limiting reactant. The yield is 0.313 mol H2O.O2

115.0 g O22¢ 1 mol

32.00 g≤ ¢6 mol H2O

2 mol O2 ≤ = 0.313 mol H2O

115.0 g C3H62¢ 1 mol

42.08 g≤ ¢ 6 mol H2O

2 mol C3H6≤ = 1.07 mol H2O

H2O15.0 g O215.0 g C3H6

C3H6 + 9 O2 ¡ 6 CO2 + 6 H2O

4.0 mol O26.0 mol CO2, 8.0 H2O,

(2.0 mol C3H8)¢ 4 mol H2O

1 mol C3H8≤ = 8.0 mol H2O produced

12.0 mol C3H82¢ 3 mol CO2

1 mol C3H8≤ = 6.0 mol CO2 produced

4.0 mol O2

C3H810 mol O2.2 mol C3H8

14.0 mol O22.0 mol C3H8

1.36 moles CO2.C3H8

180.0 g O22a 1 mol

32.00 gb ¢3 mol CO2

5 mol O2≤ = 1.50 moles CO2

120.0 g C3H82a 1 mol

44.09 gb ¢ 3 mol CO2

1 mol C3H8≤ = 1.36 moles CO2

CO2

80.0 g O220.0 g C3H8

0.375 moles CO2.O2

120.0 g O22a 1 mol

32.00 gb ¢3 mol CO2

5 mol O2≤ = 0.375 moles CO2

120.0 g C3H82a 1 mol

44.09 gb ¢ 3 mol CO2

1 mol C3H8≤ = 1.36 moles CO2

- Chapter 9 -

- 105 -


(b) Reaction between Convert each amount to moles of

is the limiting reactant. The yield is 0.521 mol H2O.

(c) Reaction between and .Convert each to moles of

Since is the limiting reactant. C3H6 will be left unreacted.

27.

The conversion is:

Using the periodic table we find that the element with is sulfur.

28.

The conversion is:

Using the periodic table we find that the element with atomic mass 65.4 is zinc.

78.5 g

1.20 mol= 65.4 g>mol

12.42 g H22a 1 mol

2.016 gb ¢ 1 mol X

1 mol H2≤ = 1.20 mol X 78.5 g X = 1.20 molX

g H2 ¡ mol H2 ¡ mol X

X + 2 HCl ¡ XCl2 + H2

32.0 g>mol

molar mass X =256

g

mol

8= 32.0

g

mol

80.0 g

0.3125 mol= 256 g>mol X8

(120.0 g O22a 1 mol

32.00 gb ¢ 1 mol X8

12 mol O2≤ = 0.3125 mol X8 80.0 g X8 = 0.3125 mol X8

g O2 ¡ mol O2 ¡ mol X8

X8 + 12 O2 ¡ 8 XO3

O2

(15.0 mol O2)¢6 mol CO2

9 mol O2≤ = 10 mol CO2

(5.0 mol C3H6)¢ 6 mol CO2

2 mol C3H6≤ = 15 mol CO2

CO2

15.0 mol of O25.0 mol of C3H6

O2

125.0 g O22¢ 1 mol

32.00 g≤ ¢6 mol H2O

9 mol O2≤ = 0.521 mol H2O

112.0 g C3H62¢ 1 mol

42.08 g≤ ¢ 6 mol H2O

2 mol C3H6≤ = 0.856 mol H2O

H2O12.0 g of C3H6 and 25.0 g of O2.

- 106 -

- Chapter 9 -


29. Limiting reactant calculation and percentage yield

Reaction between 25.0 g Al and Calculate the grams of from each reactant.

is limiting; is the theoretical yield of product.

30. Percent yield calculation


Calculate the g C needed to produce taking into account that the yield of is 86.0%. First calculate the theoretical yield of

Now calculate the grams of coke needed to produce


First calculate the grams of pure in the sample from the amount of produced.

10.540 mol C2H22¢1 mol CaC2

1 mol C2H2≤ a64.10 g CaC2

mol CaC2b = 34.6 g of pure CaC2 in the sample

C2H2CaC2

CaC2 + 2 H2O ¡ C2H2 + Ca1OH22

11.1 * 103 g CS22a 1 mol

76.15 gb ¢ 3 mol C

1 mol CS2≤ a 12.01 g

molb = 5.2 * 102 g C

1.1 * 103 g CS2.

950 g CS2

0.860= 1.1 * 103 g CS2 1theoretical yield2

CS2.CS2950 g CS2

3 C + 2 SO2 ¡ CS2 + 2 CO2

% yield = a actual yield

theoretical yieldb11002 = a 151 g

159 gb11002 = 95.0% yield of Cu

1400. g CuSO42a 1 mol

159.6 gb ¢ 1 mol Cu

1 mol CuSO4≤ a63.55 g

molb = 159 g Cu (theoretical yield)

Fe(s) + CuSO4(aq) ¡ Cu(s) + FeSO4(aq)

Percent yield = a actual yield

theoretical yieldb11002 = a 64.2 g

111 gb11002 = 57.8%

111 g AlBr3Br2

(100. g Br2)a 1 mol

159.8 gb ¢2 mol AlBr3

3 mol Br2≤ a 266.7 g

molb = 111 g AlBr3

125.0 g Al2a 1 mol

26.98 gb ¢2 mol AlBr3

2 mol Al≤ a 266.7 g

molb = 247 g AlBr3

AlBr3

100. g Br2

2 Al + 3 Br2 ¡ 2 AlBr3

- Chapter 9 -

- 107 -


Now calculate the percent in the impure sample.

33. No. There are not enough screwdrivers, wrenches or pliers. 2400 screwdrivers,3600 wrenches and 1200 pliers are needed for 600 tool sets.

34. A subscript is used to indicate the number of atoms in a formula. It cannot be changedwithout changing the identity of the substance. Coefficients are used only to balance atoms inchemical equations. They may be changed as needed to achieve a balanced equation.

35. Consider the reaction and assume that you have 1 gram of A. This does notguarantee that you will produce 1 gram of B because A and B have different molarmasses. One gram of A does not contain the same number of molecules as 1 gram of B.However, 1 mole of A does have the same number of molecules as one mole of B.(Remember, molecules always.) If you determine the number ofmoles in one gram of A and multiply by 2 to get the number of moles of B then fromthat you can determine the grams of B using its molar mass. Equations are written interms of moles not grams.

36.

(a)

(b) The conversion is:

37. (a)

Alternate Solution:

(b) 1380 g C2H5OH2a 1 mL

0.79 gb = 480 mL C2H5OH

C2H5OH = 370 g CO2 by the conservation of mass method.

750 g C2H6O6 - 380 g

1750 g C6H12O62a 1 mol

180.2 gb ¢ 2 mol CO2

1 mol C6H12O6≤ a 44.01 g

molb = 370 g CO2

1750 g C6H12O62a 1 mol

180.2 gb ¢ 2 mol C2H5OH

1 mol C6H12O6≤ a 46.07 g

molb = 380 g C2H5OH

¢0.85 g CO2

min≤ a 1 mol

44.01 gb ¢ 3 mol O2

4 mol CO2≤ a32.00 g

molb a 60.0 min

1.0 hrb =

28 g O2

hr

g CO2

min¡

mol CO2

min¡

mol O2

min¡

g O2

min¡

g O2

hr

¢0.019 mol KO2

minb110.0 min2 = 0.19 mol KO2

¢0.85 g CO2

min≤ a 1 mol

44.01 gb ¢4 mol KO2

4 mol CO2≤ =

0.019 mol KO2

min

4 KO2 + 2 H2O + 4 CO2 ¡ 4 KHCO3 + 3 O2

Á1 mole = 6.022 * 1023

A ¡ 2B

¢ 34.6 g CaC2

44.5 g sample≤11002 = 77.8% CaC2 in the impure sample

CaC2

- 108 -

- Chapter 9 -


38.

The conversion is:

39. The balanced equation is The conversion is:

(a)

(b)

(c)

(d) Reaction between 750 g of and 125 g of .Convert each amount to grams of H2O.

(e) Since is the limiting reactant, is in excess.

750 g N2H4 given - 16.8 g N2 H4 used = 730 g N2 H4 remaining

1125 g H2O22a 1 mol

34.02 gb ¢1 mol N2H4

7 mol H2O2≤ a32.05 g

molb = 16.8 g N2H4 reacted

N2H4H2O2

75.7 g H2O can be produced.

1125 g H2O22a 1 mol

34.02 gb ¢ 8 mol H2O

7 mol H2O2≤ a18.02 g

molb = 75.7 g H2O

1750 g N2H42a 1 mol

32.05 gb ¢ 8 mol H2O

1 mol N2H4≤ a18.02 g

molb = 3.4 * 103 g H2O

H2O2N2H2

1725 g H2O22a 1 mol

34.02 gb ¢1 mol N2H4

7 mol H2O2≤ a32.05 g

molb = 97.6 g N2H4

= 2.1 * 105 g H2O

1250 L H2O22a1000 mL

1 Lb ¢ 1.41 g

1 mL≤ ¢ 1 mol

34.02 g≤ a 8 mol H2O

7 mol H2O2b a 18.02 g

molb

175 kg N2H42a 1000 g

1 kgb ¢ 1 mol

32.05 g≤ ¢ 2 mol HNO3

1 mol N2H4≤ a 63.02 g

molb = 2.9 * 105 g HNO3

7 H2O2 + N2 H4 ¡ 2 HNO3 + 8 H2O

160.0 mL CH3OH2a0.72 g

mLb ¢ 1 mol

32.04 g≤ ¢ 3 mol O2

2 mol CH3OH≤ a 32.00 g

molb = 65 g O2

mL CH3OH ¡ g CH3OH ¡ mol CH3OH ¡ mol O2 ¡ g O2

2 CH3OH + 3 O2 ¡ 2 CO2 + 4 H2O

- Chapter 9 -

- 109 -



(a) Reaction between 25 g and 85 g HCl. Convert each to moles of MnCl2.

is the limiting reactant; 0.16 mol MnCl2 produced.

(b)

(c)

Theoretical yield is 91 g Cl2; Percent yield:

(d) Reaction between 25 g and 25 g KMnO4. Convert each amount to grams of CL2.

is the limiting; KMnO4 is in excess; 15 g Cl2 will be produced.

(e) Calculate the mass of unreacted

.

Unreacted


(a)

10.14 g H2S2a 1 mol

34.09 gb a2 mol Ag2S

2 mol H2Sb a 247.9 g

molb = 1.0 g Ag2S

11.1 g Ag2a 1 mol


4 mol Agb a 247.9 g

molb = 1.3 g Ag2S

4Ag + 2 H2S + O2 ¡ 2 Ag2S + 2 H2O

KMnO4 = 25 g -14 g = 11 g KMnO4 remain unreacted.

125 g HCl2a 1 mol

36.46 gb a2 mol KMnO4

16 mol HClb a158.04 g

molb = 14 g KMnO4 will react

KMnO4:

HCl

125 g KMnO42a 1 mol

158.04 gb a 5 mol Cl2

2 mol KMnO4b a 70.90 g

molb = 28 g Cl2

125 g HCl2a 1 mol

36.46 gb a 5 mol Cl2


molb = 15 g Cl2

HCl

a75 g

91 gb(100) = 82% yield

1150 g HCl2a 1 mol

36.46 gb a 5 mol Cl2


molb = 91 g Cl2

175 g KCl2a 1 mol

74.55 gb a8 mol H2O

2 mol KClb a18.02 g

molb = 73 g H2O

KMnO4

185 g HCl2a 1 mol

36.46 gb a2 mol MnCl2

16 mol HClb = 0.29 mol MnCl2

125 g KMnO42a 1 mol KMnO4

158.04 g KMnO4b a 2 mol MnCl2

2 mol KMnO4 b = 0.16 mol MnCl2

KMnO4

16 HCl + 2 KMnO4 ¡ 5 Cl2 + 2 KCl + 2 MnCl2 + 8 H2O

- 110 -

- Chapter 9 -


is limiting 1.0 g forms.

(b)

needed to completely react Ag.

42. Mass of the beaker

Since the reaction is a mole, the amount of CaO in the beaker is Convert to grams.

43.

(a) The solid is lead (II) iodide,

(b) Double displacement reaction.

(c) Calculate the moles of each reactant.

Stoichiometric quantities of reactants are used.

Theoretical yield of is 0.045 mol.

Actual yield:

Percent yield: a0.0145 mol

0.045 molb11002 = 32% yield

16.68 g PbI22a 1 mol

461.1 gb = 0.0145 mol PbI2

PbI2

115 g KI2a 1 mol

166.0 gb = 0.090 mol KI

[15 g Pb(NO3)2] + a 1 mol

331.2 gb = 0.045 mol Pb(NO3)2

PbI2 .

Pb(NO3)2(aq) + 2 KI(aq) ¡ PbI2(s) + 2 KNO3(aq)

26.095 g

-1.26 g

24.835 g

beaker + CaO

CaO

mass of the beaker

12.25 * 10-2 mol CaO2a56.08 g CaO

molb = 1.26 g CaO in the beaker.

2.25 * 10-2 mol.1:1

10.405 g H2O2a 1 mol

18.02 gb = 2.25 * 10-2 mol H2O absorbed

26.500 g

26.095 g

0.405 g

beaker + Ca(OH)2

beaker + CaO

H2O absorbed

0.17 g - 0.14 g = 0.03 grams more H2S

11.1 g Ag2a 1 mol

107.9 gb a2 mol H2S

4 mol Agb a 34.09 g

molb = 0.17 g H2S reacts

Ag2SH2S

10.080 g O22a 1 mol


1 mol O2b a 247.9 g

molb = 1.2 g Ag2S

- Chapter 9 -

- 111 -


44. Composition of a mixture of and KCl. In the mixture only KCl reacts with


(a)

(b)

(c)

needed to react wih the 180.0 g Zn

46.

2.0 mol 3.0 mol

(a) 2.0 mol Fe react with 2.0 mol to yield 2.0 mol Cu and 2.0 mol 1.0 mol is unreacted. At the completion of the reaction, there will be2.0 mol Cu, 2.0 mol and 1.0 mol

(b) Determine which reactant is limiting and then calculate the g produced fromthat reactant.

Since produces less Cu, it is the limiting reactant. Determine the mass ofproduced from 40.0 g CuSO4.FeSO4

CuSO4

140.0 g CuSO42a 1 mol

159.6 gb a 1 mol Cu

1 mol CuSO4b a 63.55 g

molb = 15.9 g Cu

120.0 g Fe2a 1 mol

55.85 gb a1 mol Cu

1 mol Feb a 63.55 g

molb = 22.8 g Cu

FeSO4

CuSO4.FeSO4,CuSO4

FeSO4.CuSO4

FeSO4(aq)+Cu(s)¡CuSO4(aq)+Fe(s)

201 g - 162 g = 39 g more HCl

1180.0 g Zn2a 1 mol

65.39 gb a2 mol HCl

1 mol Znb a36.46 g

molb = 201 g HCl reacts

1145 g Zn2a 1 mol

65.39 gb a2 mol HCl

1 mol Znb a36.46 g

molb = 162 g HCl reacted

1145 g Zn2a 1 mol

65.39 gb a 1 mol H2

1 mol Znb a2.016 g

molb = 4.47 g H2 produced

180.0 g Zn - 35 g Zn = 145 g Zn reacted with HCl

Zn + 2 HCl ¡ ZnCl2 + H2

a 7.75 g KNO3

10.00 g mixtureb11002 = 77.5% KNO3

a 2.25 g KCl

10.00 g mixtureb11002 = 22.5% KCl

10.00 g mixture - 2.25 g KCl = 7.75 g KNO3

14.33 g AgCl2a 74.55 g KCl

143.7 g AgClb = 2.25 g KCl in the mixture

KCl(aq) + AgNO3(aq) ¡ AgCl(s) + KNO3(aq)AgNO3.

KNO3

- 112 -

- Chapter 9 -


Calculate the mass of unreacted Fe.

Therefore, at the completion of thereaction, 15.9 g Cu, 38.1 g 6.0 g Fe, and no remain.

47. Limiting reactant calculation

Reaction between 40.0 g CO and 10.0 g determine the limiting reactant bycalculating the amount of that would be formed from each reactant.

CO is limiting; is in excess; 45.8 g will be produced. Calculate the mass of unreacted


(a) First calculate the theoretical yield.

Then take 84.6% of the theoretical yield to obtain the actual yield.

= 3.2 * 102 g C2H5OH

actual yield =1theoretical yield2184.62

100=13.8 * 102 g C2H5OH2184.62

100

= 3.8 * 102g C2H5OH (theoretical yield)

1750 g C6H12O62a 1 mol

180.2 gb ¢ 2 mol C2H5OH

1 mol C6H12O6≤ a 46.07 g

molb

C6H12O6 ¡ 2 C2H5OH + 2 CO2

10.0 g H2 - 5.76 g H2 = 4.2 g H2 remain unreacted

140.0 g CO2a 1 mol

28.01 gb ¢ 2 mol H2

1 mol CO≤ a 2.016 g

molb = 5.76 g H2 react

H2 :CH3OHH2

110.0 g H22a 1 mol

2.016 gb ¢1 mol CH3OH

2 mol H2≤ a32.04 g

molb = 79.5 g CH3OH

140.0 g CO2a 1 mol

28.01 gb ¢1 mol CH3OH

1 mol CO≤ a32.04 g

molb = 45.8 g CH3OH

CH3OHH2 :

CO(g) + 2 H2(g) ¡ CH3OH(l)

CuSO4FeSO4,Unreacted Fe = 20.0 g - 14.0 g = 6.0 g.


159.6 gb ¢ 1 mol Fe

1 mol CuSO4≤ a 55.85 g

molb = 14.0 g Fe will react


159.6 gb ¢ 1 mol FeSO4

1 mol CuSO4≤ a 151.9 g

molb = 38.1 g FeSO4 produced

- Chapter 9 -

- 113 -


(b) 475 g represents 84.6% of the theoretical yield. Calculate the theoretical yield.

Now calculate the g needed to produce 561 g

49. The balanced equations are:

1 mol of each salt will produce the same amount (2 mol) of AgCl. has a higherpercentage of Cl than because Mg has a lower atomic mass than Ca. Therefore, onan equal mass basis, will produce more AgCl than will

Calculations show that 1.00 g produces 3.01 g AgCl, and 1.00 g produces2.56 g AgCl.


The conversion is:


First calculate the g HCl to be produced

Then calculate the g required to produce the HCl

Finally, calculate the (96%)

11.36 * 104 g H2SO42¢1.00 g H2SO4 solution

0.96 g H2SO4≤ a 1 kg

1000 gb = 14 kg concentrated H2SO4

kg H2SO4

11.01 * 104 g HCl2a 1 mol

36.46 gb ¢ 1 mol H2SO4

2 mol HCl≤ a 98.09 g

1 molb = 1.36 * 104 g H2SO4

H2SO4

120.0 L HCl solution2a1000 mL

1 Lb a 1.20 g

1.00 mLb10.4202 = 1.01 * 104 g HCl

H2SO4 + 2 NaCl ¡ Na2SO4 + 2 HCl

¢ 4.1 kg Li2O

astronaut day≤130 days213 astronauts2 = 3.7 * 102 kg Li2O

¢ 2500 g H2O

astronaut day≤ a 1 mol

18.02 gb ¢1 mol Li2O

1 mol H2O≤ a 29.88 g

molb a 1 kg

1000 gb =

4.1 kg Li2O

astronaut day

g H2O ¡ mol H2O ¡ mol Li2O ¡ g Li2O ¡ kg Li2O

Li2O + H2O ¡ 2 LiOH

CaCl2MgCl2

CaCl2 .MgCl2

CaCl2

MgCl2

MgCl2(aq) + 2 AgNO3(aq) ¡ Mg1NO322(aq) + 2 AgCl(s)

CaCl2(aq) + 2 AgNO3(aq) ¡ Ca(NO3)2(aq) + 2 AgCl(s)

1561 g C2H5OH2a 1 mol

46.07 gb ¢ 1 mol C6H12O6

2 mol C2H5OH≤ a 180.2 g

molb = 1.10 * 103 g C6H12O6

C2H5OH.C6H12O6

theoretical yield =475 g

0.846= 561 g C2H5OH

C2H5OH

- 114 -

- Chapter 9 -


52. Percent yield of

Alternate Solution:Calculation of yield. There are three chemical steps to the formation of Each stephas a 10% loss of yield.

Step 1:Step 2:Step 3:

Now calculate the grams of product. One mole of sulfur will yield a maximum of 1 molTherefore 3.118 mol S will give a maximum of 3.118 mol

53. According to the equations, the moles of come from both reactions and the molescome from only the first reaction.

So the mol

in the sample110.00 g NaHCO3 + Na2CO32 - 6.00 g NaHCO3 = 4.00 g Na2CO3

10.0714 mol NaHCO32a 84.01 g

molb = 6.00 g NaHCO3 in the sample

NaHCO3 = 2 * mol H2O = 2 * 0.0357 mol = 0.0714 mol NaHCO3

H2OCO2

13.118 mol S2a1 mol H2SO4

1 mol Sb a98.09 g

molb10.72902 = 223.0 g H2SO4 yield

H2SO4.H2SO4.

81.00% yield - 10% = 72.90% yield 90.00% yield - 10% = 81.00% yield

100% yield - 10% = 90.00% yield

H2SO4.

¢2.273 mol H2SO4

3.118 mol H2SO4≤11002 = 72.90% yield

13.118 mol S2a1 mol H2SO4

1 mol Sb = 3.118 mol H2SO4 (theoretical yield)

12.273 mol H2SO42a98.09 g

molb = 223.0 g H2SO4 formed

-0.2525

2.2725

2.525 mol SO3 ¡ 2.525 mol H2SO4 - 10% = 2.273 mol H2SO4

-0.2806

2.5254

2.806 mol SO2 ¡ 2.806 mol SO3 - 10% = 2.525 mol SO3

-0.3118

2.8062

3.118 mol S ¡ 3.118 mol SO2 - 10% = 2.806 mol SO2

1100.0 g S2a 1 mol

32.07 gb = 3.118 mol S to start with

H2SO4

- Chapter 9 -

- 115 -



lost by heating

Because the lost came only from we can use it to calculate the amount ofin the mixture.

The conversion is:


The conversion is:

Now calculate the number of 400. mg tablets that can be made from

56.

In the first reaction:

This is a ratio of

Therefore, P is the limiting reactant and the produced is:P4O10

0.646 mol P

0.938 mol O2=

3.44 mol P

5.00 mol O2

130.0 g O22a 1 mol

32.00 gb = 0.938 mol O2

120.0 g P2a 1 mol

30.97 gb = 0.646 mol P

P4O10 + 6 H2O ¡ 4 H3PO4

4 P + 5 O2 ¡ P4O10

¢5.3 g Al(OH)3

day≤ a1000 mg

gb a 1 tablet

400. mgb = 13 tablets>day

5.3 g Al(OH)3

a2.5 L

dayb a3.0 g HCl

Lb a 1 mol

36.46 gb ¢1 mol Al1OH23

3 mol HCl≤ a78.00 g

molb = 5.3 g Al1OH23>day

L HCl ¡ g HCl ¡ mol HCl ¡ mol Al(OH)3 : g Al(OH)3

Al(OH)3(s) + 3 HCl(aq) ¡ AlCl3(aq) + 3 H2O(l)

a 8.61 g KClO3

12.82 g sampleb11002 = 67.2% KClO3

13.37 g O22a 1 mol

32.00 gb a2 mol KClO3

3 mol O2b a122.6 g

molb = 8.61 g KClO3 in the mixture

g O2 ¡ mol O2 ¡ mol KClO3 ¡ g KClO3

KClO3

KClO3,O2

12.82 g mixture - 9.45 g residue = 3.37 g O2

2 KClO3 ¡ 2 KCl + 3 O2

a4.00 g Na2CO3

10.00 gb11002 = 40.0% Na2CO3

a6.00 g NaHCO3

10.00 gb11002 = 60.0% NaHCO3

- 116 -

- Chapter 9 -


In the second reaction:

and we have The ratio of is

Therefore, is the limiting reactant and the produced is:

10.832 mol H2O2¢4 mol H3PO4

6 mol H2O≤ a 97.99 g

molb = 54.4 g H3PO4

H3PO4H2O

0.832 mol

0.162 mol=

5.14 mol

1.00 mol

H2O

P4O100.162 mol P4O10.

115.0 g H2O2a 1 mol

18.02 gb = 0.832 mol H2O

10.646 mol P2¢1 mol P4O10

4 mol P≤ = 0.162 mol P4O10

- Chapter 9 -

- 117 -


chapter 9 calculations from chemical …faculty.chemeketa.edu/lemme/ch...

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