Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN: 9781337399425
Author: Steven S. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 9, Problem 13QAP

For each of the following balanced chemical equations, calculate how many grams of the product(s) would be produced by complete reaction of 0.125 mole of the first reactant.

msp;  AgNO 3 ( a q ) + LiOH ( a q ) AgOH ( s ) + LiNO 3 ( a q )

msp;  Al 2 ( SO 4 ) 3 ( a q ) + 3 CaCl 2 ( a q ) 2 AlCl 3 ( a q ) + 3 CaSO 4 ( s )

msp;  CaCO 3 ( s ) + 2 HCl ( a q ) CaCl 2 ( a q ) + CO 2 ( g ) + H 2 O ( l )

msp;  2 C 4 H 10 ( g ) + 13 O 2 ( g ) 8 CO 2 ( g ) + 10 H 2 O ( g )

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

Grams of the product(s) produced by complete reaction of 0.125  mole of the first reactant in the given balanced chemical equation should be calculated.

Concept Introduction:

For a substance, number of moles is related to mass and molar mass of the substance as follows:

n=mM

Here, m is mass of substance in g and M is molar mass of substance in g/mol.

Thus, from number of moles, mass can be calculated as follows:

m=n×M.

Answer to Problem 13QAP

Mass of AgOH ( s ) Produced =  15.6 g

Mass of LiNO3 ( a q ) produced =  8.62 g.

Explanation of Solution

According to the balanced equation, mole ratio between AgNO3 ( a q ) and AgOH ( s ) = 1 mol AgOH(s)1 mol AgNO3 (aq)

So, Amount of AgOH ( s ) produced = 0.125 mol AgNO3(aq)×

1 mol AgOH(s)1 mol AgNO3 (aq)

= 0.125 mol

Molar mass of AgOH ( s ) = (107.9+16.00+1.028) g/mol = 124.928 g/ mol

Mass of AgOH ( s ) Produced = 0.125 mol ×

124.928 g/ mol

=  15.6 g

According to the balanced equation, mole ratio between AgNO3 ( a q ) and LiNO3 ( a q ) = 1 mol LiNO3 (aq)1 mol AgNO3 (aq)

So, Amount of LiNO3 ( a q ) produced = 0.125 mol AgNO3(aq)×

1 mol LiNO3 (aq)1 mol AgNO3 (aq)

= 0.125 mol

Molar mass of LiNO3 ( a q ) = {(6.941 + 14.01 + (3 × 16.00)}g/mol = 68.951 g/mol

Mass of LiNO3 ( a q ) produced = 0.125 mol ×

68.951 g/mol

=  8.62 g.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

Grams of the product(s) produced by complete reaction of 0.125  mole of the first reactant in the given balanced chemical equation should be calculated.

Concept Introduction:

For a substance, number of moles is related to mass and molar mass of the substance as follows:

n=mM

Here, m is mass of substance in g and M is molar mass of substance in g/mol.

Thus, from number of moles, mass can be calculated as follows:

m=n×M.

Answer to Problem 13QAP

Mass of AlCl3 ( a q ) produced = 33.3 g

Mass of CaSO4 ( s ) produced = 51.1 g.

Explanation of Solution

According to the balanced equation, mole ratio between Al2 (SO4 )3 ( a q ) and AlCl3 ( a q ) = 2 mol AlCl3 (aq)1 mol Al2(SO4)3 (aq)

So, Amount of AlCl3 ( a q ) produced = 0.125 mol Al2(SO4)3 (aq)×

2 mol AlCl3 (aq)1 mol Al2(SO4)3 (aq)

= 0.250 mol

Molar mass of AlCl3 ( a q ) = {26.98 + (3 × 35.45)} g/mol = 133.33 g/ mol

Mass of AlCl3 ( a q ) Produced = 0.250 mol × 133.33 g/ mol

= 33.3 g

According to the balanced equation, mole ratio between Al2 (SO4 )3 ( a q ) and CaSO4 ( s ) = 3 mol CaSO4(s)1 mol Al2(SO4)3 (aq)

So, Amount of CaSO4 ( s ) produced = 0.125 mol Al2(SO4)3 (aq)×

3 mol CaSO4(s)1 mol Al2(SO4)3 (aq)

= 0.375 mol

Molar mass of CaSO4 ( s ) = {40.08 + 32.07 + (4 × 16.00)} g/mol = 136.15 g/mol

Mass of CaSO4 ( s ) produced = 0.375 mole × 136.15 g/mol

= 51.1 g.

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

Grams of the product(s) produced by complete reaction of 0.125  mole of the first reactant in the given balanced chemical equation should be calculated.

Concept Introduction:

For a substance, number of moles is related to mass and molar mass of the substance as follows:

n=mM

Here, m is mass of substance in g and M is molar mass of substance in g/mol.

Thus, from number of moles, mass can be calculated as follows:

m=n×M.

Answer to Problem 13QAP

Mass of CaCl2 ( a q ) Produced =  13.9 g

Mass of CO2 ( g ) produced = 5.50 g

Mass of H2 O( l ) produced = 2.23 g.

Explanation of Solution

According to the balanced equation, mole ratio between CaCO3 ( s ) and CaCl2 ( a q ) = 1 mol CaCl2(aq)1 mol CaCO3 (s)

So, Amount of CaCl2 ( a q ) produced = 0.125 mol CaCO3 (s)×

1 mol CaCl2(aq)1 mol CaCO3 (s)

= 0.125 mol

Molar mass of CaCl2 ( a q ) = {40.08 + (2 × 35.45)} g/mol = 110.98 g/ mol

Mass of CaCl2 ( a q ) Produced = 0.125 mol ×

110.98 g/ mol

=  13.9 g

According to the balanced equation, mole ratio between CaCO3 ( s ) and CO2 ( g ) = 1 mol CO2(g)1 mol CaCO3 (s)

So, Amount of CO2 ( g ) produced = 0.125 mol CaCO3 (s)×

1 mol CO2(g)1 mol CaCO3 (s)

= 0.125 mol

Molar mass of CO2 ( g ) = {12.01 + (2 × 16.00)} g/mol = 44.01 g/mol

Mass of CO2 ( g ) produced = 0.125 mol ×

44.01 g/mol

= 5.50 g

According to the balanced equation, mole ratio between CaCO3 ( s ) and H2 O( l ) = 1 mol H2O(l)1 mol CaCO3 (s)

So, Amount of H2 O( l ) produced = 0.125 mol CaCO3 (s)×

1 mol H2O(l)1 mol CaCO3 (s)

= 0.125 mol

Molar mass of H2 O( l ) = {(2 × 1.008) + 16.00} g/mol = 18.016 g/mol

Mass of H2 O( l ) produced = 0.125 mol ×

18.016 g/mol

= 2.23 g.

Expert Solution
Check Mark
Interpretation Introduction

(d)

Interpretation:

Grams of the product(s) produced by complete reaction of 0.125  mole of the first reactant in the given balanced chemical equation should be calculated.

Concept Introduction:

For a substance, number of moles is related to mass and molar mass of the substance as follows:

n=mM

Here, m is mass of substance in g and M is molar mass of substance in g/mol.

Thus, from number of moles, mass can be calculated as follows:

m=n×M.

Answer to Problem 13QAP

Mass of CO2 ( g ) produced =  22.0 g

Mass of H2 O( g ) produced =  11.3 g.

Explanation of Solution

According to the balanced equation, mole ratio between C4 H1 0 ( g ) and CO2 ( g ) = 4 mol CO2(g)1 mol C4H10 (g)

So, Amount of CO2 ( g ) produced = 0.125 mol C4H10 (g)×

4 mol CO2(g)1 mol C4H10 (g)

= 0.500 mol

Molar mass of CO2 ( g ) = {12.01 + (2 × 16.00)} g/mol = 44.01 g/mol

Mass of CO2 ( g ) produced = 0.500 mol × 44.01 g/mol

=  22.0 g

According to the balanced equation, mole ratio between C4 H1 0 ( g ) and H2 O( g ) = 5 mol H2O(g)1 mol C4H10 (g)

So, Amount of H2 O( g ) produced = 0.125 mol C4H10 (g)×

5 mol H2O(g)1 mol C4H10 (g)

= 0.625 mol

Molar mass of H2 O( g ) = {(2 × 1.008) + 16.00} g/mol = 18.016 g/mol

Mass of H2 O( g ) produced = 0.625 mol × 18.016 g/mol

=  11.3 g.

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Chapter 9 Solutions

Introductory Chemistry: A Foundation

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