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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Textbook Question
Chapter 9, Problem 50QAP

For each of the following unbalanced chemical equations, suppose that exactly 15.0 g of each reactant are taken. Using Before− Change−After (BCA) tables, determine which reactant is limiting, and calculate what mass of each product is expected. (Assume that the limiting reactant is completely consumed.)

msp;  Al ( s ) + HCl ( a q ) AlCl 3 ( a q ) + H 2 ( g )

msp;  NaOH ( a q ) + CO 2 ( g ) Na 2 CO 3 ( a q ) + H 2 O ( l )

msp;  Pb ( NO 3 ) 2 ( a q ) + HCl ( a q ) PbCl 2 ( s ) + NHO 3 ( a q )

msp;  K ( s ) + I 2 ( s ) KI ( s )

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 equationN2       +         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 50QAP

The limiting reagent is HCl.

Mass of AlCl3 produce = 18.3 g

Mass of H2 produce = 0.415 g.

Explanation of Solution

Number of moles of Al = 15.0 g26.98 g/mol = 0.556 mol

Number of moles of HCl = 15.0 g36.458 g/mol = 0.411 mol

Possibility I: if Al runs out first

Balanced equation 2Al(s)      +        6HCl(aq)            2AlCl3(aq)    +      3H2(g)

Before 0.556 mol 0.411 mol 0 0

Change 0.556 mol 1.67 mol +0.556 mol +0.834 mol

_________________________________________________________________________

After 0 1.26 mol 0.556 mol 0.834 mol

Possibility II: if HCl runs out first

Balanced equation2Al(s)      +        6HCl(aq)            2AlCl3(aq)    +      3H2(g)

Before0.556 mol 0.411 mol 0 0

Change0.137 mol 0.411 mol +0.137 mol +0.206 mol

_________________________________________________________________________

After 0.419 mol 0 0.137 mol 0.206 mol

According to BCA tables, Al is not the limiting reactant as to react with all the Al we need 1.26 mol more HCl than we have. If HCl is the limiting reagent no negative results seen in the After column. So the limiting reagent is HCl.

Mass of AlCl3 produce = 0.137 mol × 133.33 g/mol = 18.3 g

Mass of H2 produce = 0.206 mol × 2.016 g/mol = 0.415 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 equationN2       +         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 50QAP

The limiting reagent is NaOH

Mass of Na2 CO3 produce = 19.9 g

Mass of H2 O produce = 3.39 g.

Explanation of Solution

Number of moles of NaOH = 15.0 g39.998 g/mol = 0.375 mol

Number of moles of CO2 = 15.0 g44.01 g/mol = 0.341 mol

Possibility I: if NaOH runs out first

Balanced equation2NaOH(aq)      +          CO2(g)                Na2CO3(aq)        +         H2O(l)

Before0.375 mol 0.341 mol 0 0

Change0.375 mol 0.188 mol +0.188 mol +0.188 mol

______________________________________________________________________________

After 0 0.153 mol 0.188 mol 0.188 mol

Possibility II: if CO2 runs out first

Balanced equation2NaOH(aq)      +          CO2(g)                Na2CO3(aq)        +         H2O(l)

Befor0.375 mol 0.341 mol 0 0

Change0.682 mol 0.341 mol +0.341 mol +0.341 mol

_________________________________________________________________________

After 0.307 mol 0 0.341 mol 0.341 mol

According to BCA tables, CO2 is not the limiting reactant as to react with all the NaOH, we need 0.307 mol more NaOH than we have. If NaOH is the limiting reagent no negative results seen in the After column. So the limiting reagent is NaOH.

Mass of Na2 CO3 produce = 0.188 mol × 105.99 g/mol = 19.9 g

Mass of H2 O produce = 0.188 mol × 18.016 g/mol = 3.39 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 equationN2       +         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 50QAP

The limiting reagent is Pb(NO3 )2 Mass of PbCl2 produce = 12.6 g

Mass of HNO3 produce = 5.71 g.

Explanation of Solution

Number of moles of Pb(NO3 )2 = 15.0 g331.22 g/mol = 0.0453 mol

Number of moles of HCl = 15.0 g36.458 g/mol = 0.411 mol

Possibility I: if Pb(NO3 )2 runs out first

Balanced equation Pb(NO3)2(aq)      +          2HCl(aq)                 PbCl2(s)        +         2HNO3(aq)

Before 0.0453 mol 0.411 mol 0 0

Change 0.0453 mol 0.0906 mol +0.0453 mol +0.0906 mol

______________________________________________________________________________

After 0 0.320 mol 0.0453 mol 0.0906 mol

Possibility II: if HCl runs out first

Balanced equation Pb(NO3)2(aq)      +          2HCl(aq)                 PbCl2(s)        +         2HNO3(aq)

Before 0.0453 mol 0.411 mol 0 0

Change 0.206 mol 0.411 mol +0.206 mol +0.411 mol

_________________________________________________________________________

After 0.161 mol 0 0.206 mol 0.411 mol

According to BCA tables, HCl is not the limiting reactant as, to react with all the HCl, we need 0.161 mol more Pb(NO3 )2 than we have. If Pb(NO3 )2 is the limiting reagent no negative results seen in the After column. So the limiting reagent is Pb(NO3 )2.

Mass of PbCl2 produce = 0.0453 mol × 278.1 g/mol = 12.6 g

Mass of HNO3 produce = 0.0906 mol × 63.018 g/mol = 5.71 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 equationN2       +         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 50QAP

The limiting reagent is I2.

Mass of KI produce = 19.6 g.

Explanation of Solution

Number of moles of K = 15.0 g39.00 g/mol = 0.385 mol

Number of moles of I2 = 15.0 g253.8 g/mol = 0.0591 mol

Possibility I: if K runs out first

Balanced equation 2K(s)      +          I2(s)                 2KI(s) 

Before 0.385 mol 0.0591 mol 0

Change 0.385 mol 0.193 mol +0.385 mol

______________________________________________________

After 0 -0.134 mol 0.385 mol

Possibility II: if I2 runs out first

Balanced equation 2K(s)      +          I2(s)                 2KI(s) 

Before 0.385 mol 0.0591 mol 0

Change 0.118 mol 0.0591 mol +0.118 mol

______________________________________________________

After 0.267mol 0 0.118 mol

According to BCA tables, K is not the limiting reactant as, to react with all the K, we need 0.134 mol more I2 than we have. If I2 is the limiting reagent no negative results seen in the After column. So the limiting reagent is I2.

Mass of KI produce = 0.118 mol × 165.9 g/mol = 19.6 g.

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Chapter 9 Solutions

Introductory Chemistry: A Foundation

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