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 88AP

For each of the following unbalanced chemical equations, suppose exactly 5.0 g of each reactant is taken. Determine which reactant is limiting, and calculate what mass of each product is expected, assuming that the limiting reactant is completely consumed.

msp;  Na ( s ) + Br 2 ( l ) NaBr ( s )

msp;  Zn ( s ) + CuSO 4 ( a q ) ZnSO 4 ( a q ) + Cu ( s )

msp;  NH 4 Cl ( a q ) + NaOH ( a q ) NH 3 ( g ) + H 2 O ( l ) + NaCl ( a q )

msp;  Fe 2 O 3 ( s ) + CO ( g ) Fe ( s ) + CO 2 ( g )

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

Using Before-Change-After (BCA) tables, the limiting reactant should be determined in the given unbalanced equation, supposing that exactly 15.0 g of each reactant are taken. And mass of each product expected should be calculated assuming that the limiting reactant is completely consumed.

Concept Introduction:

To determine how much product can be formed from a given mixture of reactants, we have to look for the reactant that is limiting; the one that runs out first and thus limits the amount of product that can form. The reactant that runs out first limiting the amount of products form is called the limiting reactant or limiting reagent.

To determine limiting reactant, first we should have a balanced equation. Then we include the information in Before-Change-After table.

E.g

Balanced equation

N2       +         3H2             2NH3

Before

Change

After

Starting amounts of reactants are presented in before row. The change row represents how much of each substance reacts or is produced. The after row represents how much of each substance remain in the final reaction mixture. The ratio of the numbers in the change row has to be the same as the ratio of the coefficients in the balanced equation.

Answer to Problem 88AP

The limiting reagent is Br2

Mass of NaBr produce = 6.44 g.

Explanation of Solution

Number of moles of Na = 5.0 g22.99 g/mol = 0.217 mol

Number of moles of Br2 = 5.0 g159.8 g/mol = 0.0313 mol

Possibility I: if Na runs out first

Balanced equation

2Na(s)      +        Br2(l)            2NaBr(s) 

Before

0.217 mol

0.0313 mol

0

Change

0.217 mol

0.109 mol

+0.217 mol

________________________________________________________

After

0

0.0777 mol

0.217 mol

Possibility II: if Br2 runs out first

Balanced equation

2Na(s)      +        Br2(l)            2NaBr(s) 

Before

0.217 mol

0.0313 mol

0

Change

0.0626 mol

0.0313 mol

+0.0626 mol

________________________________________________________

After

0.154 mol

0

0.0626 mol

According to BCA tables, Na is not the limiting reactant as to react with all the Na we need 0.0777 mol more Br2 than we have. If Br2 is the limiting reagent no negative results seen in the After column. So the limiting reagent is Br2

Mass of NaBr produce = 0.0626 mol × 102.89 g/mol = 6.44 g.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

Using Before-Change-After (BCA) tables, the limiting reactant should be determined in the given unbalanced equation, supposing that exactly 15.0 g of each reactant are taken. And mass of each product expected should be calculated assuming that the limiting reactant is completely consumed.

Concept Introduction:

To determine how much product can be formed from a given mixture of reactants, we have to look for the reactant that is limiting; the one that runs out first and thus limits the amount of product that can form. The reactant that runs out first limiting the amount of products form is called the limiting reactant or limiting reagent.

To determine limiting reactant, first we should have a balanced equation. Then we include the information in Before-Change-After table.

E.g

Balanced equation

N2       +         3H2             2NH3

Before

Change

After

Starting amounts of reactants are presented in before row. The change row represents how much of each substance reacts or is produced. The after row represents how much of each substance remain in the final reaction mixture. The ratio of the numbers in the change row has to be the same as the ratio of the coefficients in the balanced equation.

Answer to Problem 88AP

The limiting reagent is CuSO4 Mass of ZnSO4 produce = 5.05 g

Mass of Cu produce = 1.99 g.

Explanation of Solution

Number of moles of Zn = 5.0 g65.38 g/mol = 0.0765 mol

Number of moles of CuSO4 = 5.0 g159.62 g/mol = 0.0313 mol

Possibility I: if Zn runs out first

Balanced equation

Zn(s)      +          CuSO4(aq)                ZnSO4(aq)        +         Cu(s)

Before

0.0765 mol

0.0313 mol

0

0

Change

0.0765 mol

0.0765 mol

+0.0765 mol

+0.0765 mol

______________________________________________________________________________

After

0 -0.0452 mol

0.0765 mol

0.0765 mol

Possibility II: if CuSO4 runs out first

Balanced equation

Zn(s)      +          CuSO4(aq)                ZnSO4(aq)        +         Cu(s)

Before

0.0765 mol

0.0313 mol

0

0

Change

0.0313 mol

0.0313 mol

+0.0313 mol

+0.0313 mol

______________________________________________________________________________

After

0.0452 mol

0

0.0313 mol

0.0313 mol

According to BCA tables, Zn is not the limiting reactant as to react with all the Zn, we need 0.0452 mol more CuSO4 than we have. If CuSO4 is the limiting reagent no negative results seen in the After column. So the limiting reagent is CuSO4

Mass of ZnSO4 produce = 0.0313 mol × 161.45 g/mol = 5.05 g

Mass of Cu produce = 0.0313 mol × 63.55 g/mol = 1.99 g.

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

Using Before-Change-After (BCA) tables, the limiting reactant should be determined in the given unbalanced equation, supposing that exactly 15.0 g of each reactant are taken. And mass of each product expected should be calculated assuming that the limiting reactant is completely consumed.

Concept Introduction:

To determine how much product can be formed from a given mixture of reactants, we have to look for the reactant that is limiting; the one that runs out first and thus limits the amount of product that can form. The reactant that runs out first limiting the amount of products form is called the limiting reactant or limiting reagent.

To determine limiting reactant, first we should have a balanced equation. Then we include the information in Before-Change-After table.

E.g

Balanced equation

N2       +         3H2             2NH3

Before

Change

After

Starting amounts of reactants are presented in before row. The change row represents how much of each substance reacts or is produced. The after row represents how much of each substance remain in the final reaction mixture. The ratio of the numbers in the change row has to be the same as the ratio of the coefficients in the balanced equation.

Answer to Problem 88AP

The limiting reagent is NH4 Cl

Mass of H2 O produce = 1.68 g

Mass of NH3 produce = 1.59 g

Mass of NaCl produce = 5.46 g.

Explanation of Solution

Number of moles of NH4 Cl = 5.0 g53.492 g/mol = 0.0935 mol

Number of moles of NaOH = 5.0 g39.998 g/mol = 0.125 mol

Possibility I: if NH4 Cl runs out first

Balanced equation

NH4Cl(aq)    +        NaOH(aq)            H2O(l)    +      NH3(g)  +     NaCl(aq)

Before

0.0935 mol

0.125 mol

0

0 0

Change

0.0935 mol

0.0935 mol +0.0935 mol

+0.0935 mol

+0.0935 mol

______________________________________________________________________________

After

0

0.116 mol

0.0935 mol

0.0935 mol

0.0935 mol

Possibility II: if NaOH runs out first

Balanced equation

NH4Cl(aq)    +        NaOH(aq)            H2O(l)    +      NH3(g)  +     NaCl(aq)

Before

0.0935 mol

0.125 mol

0

0 0

Change

0.125 mol

0.125 mol +0.125 mol

+0.125 mol

+0.125 mol

______________________________________________________________________________

After

0.0315 mol

0

0.125 mol

0.125 mol

0.125 mol

According to BCA tables, NaOH is not the limiting reactant as, to react with all the NaOH, we need 0.0315 mol more NH4 Cl than we have. If NH4 Cl is the limiting reagent no negative results seen in the After column. So the limiting reagent is NH4 Cl.

Mass of H2 O produce = 0.0935 mol × 18.016 g/mol = 1.68 g

Mass of NH3 produce = 0.0935 mol × 17.034 g/mol = 1.59 g

Mass of NaCl produce = 0.0935 mol × 58.44 g/mol = 5.46 g.

Expert Solution
Check Mark
Interpretation Introduction

(d)

Interpretation:

Using Before-Change-After (BCA) tables, the limiting reactant should be determined in the given unbalanced equation, supposing that exactly 15.0 g of each reactant are taken. And mass of each product expected should be calculated assuming that the limiting reactant is completely consumed.

Concept Introduction:

To determine how much product can be formed from a given mixture of reactants, we have to look for the reactant that is limiting; the one that runs out first and thus limits the amount of product that can form. The reactant that runs out first limiting the amount of products form is called the limiting reactant or limiting reagent.

To determine limiting reactant, first we should have a balanced equation. Then we include the information in Before-Change-After table.

E.g

Balanced equation

N2       +         3H2             2NH3

Before

Change

After

Starting amounts of reactants are presented in before row. The change row represents how much of each substance reacts or is produced. The after row represents how much of each substance remain in the final reaction mixture. The ratio of the numbers in the change row has to be the same as the ratio of the coefficients in the balanced equation.

Answer to Problem 88AP

The limiting reagent is Fe2 O3

Mass of Fe produce = 6.65 g

Mass of CO2 produce = 7.88 g.

Explanation of Solution

Number of moles of Fe2 O3 = 5.0 g159.7 g/mol = 0.0313 mol

Number of moles of CO = 5.0 g28.01 g/mol = 0.179 mol

Possibility I: if Fe2 O3 runs out first

Balanced equation

Fe2O3(s)      +          3CO(g)                 2Fe(s)       +         3CO2(g) 

Before

0.0313 mol 0.179 mol

0

0

Change

0.0313 mol 0.0939 mol

+0.626 mol

+0.0939 mol

____________________________________________________________________________

After

0 0.0851 mol

0.626 mol

0.626 mol

Possibility II: if CO runs out first

Balanced equation

Fe2O3(s)      +          3CO(g)                 2Fe(s)       +         3CO2(g) 

Before

0.0313 mol 0.179 mol

0

0

Change

0.0597 mol 0.179 mol

+0.119 mol

+0.179 mol

____________________________________________________________________________

After

0.0284 mol

0

0.119 mol

0.179 mol

According to BCA tables, CO is not the limiting reactant as, to react with all the CO, we need 0.134 mol more Fe2 O3 than we have. If Fe2 O3 is the limiting reagent no negative results seen in the After column. So the limiting reagent is Fe2 O3.

Mass of Fe produce = 0.119 mol × 55.85 g/mol = 6.65 g

Mass of CO2 produce = 0.179 mol × 44.01 g/mol = 7.88 g.

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Chapter 9 Solutions

Introductory Chemistry: A Foundation

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