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;  $\text{Na}\left(s\right)+{\text{Br}}_2\left(l\right)\to \text{NaBr}\left(s\right)$

msp;  $\text{Zn}\left(s\right)+{\text{CuSO}}_4\left(aq\right)\to {\text{ZnSO}}_4\left(aq\right)+\text{Cu}\left(s\right)$

msp;  ${\text{NH}}_4\text{Cl}\left(aq\right)+\text{NaOH}\left(aq\right)\to {\text{NH}}_3\left(g\right)+{\text{H}}_2\text{O}\left(l\right)+\text{NaCl}\left(aq\right)$

msp;  ${\text{Fe}}_2{\text{O}}_3\left(s\right)+\text{CO}\left(g\right)\to \text{Fe}\left(s\right)+{\text{CO}}_2\left(g\right)$

Interpretation Introduction

(a)

Interpretation:

Using Before-Change-After (BCA) tables, the limiting reactant should be determined in the given unbalanced equation, supposing that exactly $15.0\text{ g}$ 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

${\text{N}}_{\text{2}} + 3{\text{H}}_{\text{2}} \to 2{\mathrm{NH}}_3$

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 = $\text{6}\text{.44 g}$.

Explanation of Solution

Number of moles of Na = $\frac{5.0\text{ g}}{22.99\text{ g/mol}}$ = $0.217\text{ mol}$

Number of moles of Br₂ = $\frac{5.0\text{ g}}{159.8\text{ g/mol}}$ = $0.0313\text{ mol}$

Possibility I: if Na runs out first

Balanced equation

$2{\text{Na}}_{(s)}+{\text{ Br}}_2{}_{(l)}\to 2{\text{NaBr}}_{(s)}$

Before

$0.217\text{ mol}$

$0.0313\text{ mol}$

$0$

Change

$-0.217\text{ mol}$

$-0.109\text{ mol}$

$+0.217\text{ mol}$

________________________________________________________

After

$0$

$-0.0777\text{m}\text{o}\text{l}$

$0.217\text{m}\text{o}\text{l}$

Possibility II: if Br₂ runs out first

Balanced equation

$2{\text{Na}}_{(s)}+{\text{ Br}}_2{}_{(l)}\to 2{\text{NaBr}}_{(s)}$

Before

$0.217\text{ mol}$

$0.0313\text{ mol}$

$0$

Change

$-0.062\text{6 mol}$

$-0.0313\text{ mol}$

$+0.0626\text{ mol}$

________________________________________________________

After

$0.15\text{4 mol}$

$0$

$0.0626\text{ mol}$

According to BCA tables, Na is not the limiting reactant as to react with all the Na we need $\text{0}\text{.0777 mol}$ 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 = $0.06\text{26 mol}$ × $102.89\text{ g/mol}$ = $\text{6}\text{.44 g}$.

Interpretation Introduction

(b)

Interpretation:

Using Before-Change-After (BCA) tables, the limiting reactant should be determined in the given unbalanced equation, supposing that exactly $15.0\text{ g}$ 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

${\text{N}}_{\text{2}} + 3{\text{H}}_{\text{2}} \to 2{\text{NH}}_{\text{3}}$

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 = $\text{5}\text{.05 g}$

Mass of Cu produce = $\text{1}\text{.99 g}$.

Explanation of Solution

Number of moles of Zn = $\frac{5.0\text{ g}}{\text{65}\text{.38 g/mol}}$ = $0.07\text{65 mol}$

Number of moles of CuSO₄ = $\frac{5.0\text{ g}}{\text{159}\text{.62 g/mol}}$ = $0.0313\text{ mol}$

Possibility I: if Zn runs out first

Balanced equation

${\text{Zn}}_{(s)}{\text{ + CuSO}}_{4(aq)}\to {\text{ ZnSO}}_{4(aq)}+{\text{ Cu}}_{(s)}$

Before

$0.076\text{5 mol}$

$0.0313\text{ mol}$

$0$

$0$

Change

$-0.076\text{5 mol}$

$-0.076\text{5 mol}$

$+0.076\text{5 mol}$

$+0.07\text{65 mol}$

______________________________________________________________________________

After

$0$ $\text{-0}\text{.0452 mol}$

$0.0765\text{m}\text{o}\text{l}$

$0.076\text{5 mol}$

Possibility II: if CuSO₄ runs out first

Balanced equation

${\text{Zn}}_{(s)}{\text{ + CuSO}}_{4(aq)}\to {\text{ ZnSO}}_{4(aq)}+{\text{ Cu}}_{(s)}$

Before

$0.076\text{5 mol}$

$0.0313\text{ mol}$

$0$

$0$

Change

$-0.0313\text{ mol}$

$-0.0313\text{ mol}$

$+0.0313\text{ mol}$

$+0.0313\text{ mol}$

______________________________________________________________________________

After

$0.045\text{2 mol}$

$0$

$0.0313\text{ mol}$

$0.0313\text{ mol}$

According to BCA tables, Zn is not the limiting reactant as to react with all the Zn, we need $\text{0}\text{.0452 mol}$ 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 = $0.0313\text{ mol}$ × $161.45\text{ g/mol}$ = $\text{5}\text{.05 g}$

Mass of Cu produce = $0.0313\text{ mol}$ × $\text{63}\text{.55 g/mol}$ = $\text{1}\text{.99 g}$.

Interpretation Introduction

(c)

Interpretation:

Using Before-Change-After (BCA) tables, the limiting reactant should be determined in the given unbalanced equation, supposing that exactly $15.0\text{ g}$ 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

${\text{N}}_{\text{2}} + {\text{ 3H}}_{\text{2}} \to {\text{2NH}}_{\text{3}}$

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 = $\text{1}\text{.68 g}$

Mass of NH₃ produce = $\text{1}\text{.59 g}$

Mass of NaCl produce = $\text{5}\text{.46 g}$.

Explanation of Solution

Number of moles of NH₄ Cl = $\frac{5.0\text{ g}}{\text{53}\text{.492 g/mol}}$ = $0.093\text{5 mol}$

Number of moles of NaOH = $\frac{5.0\text{ g}}{39.998\text{ g/mol}}$ = $0.125\text{ mol}$

Possibility I: if NH₄ Cl runs out first

Balanced equation

${\text{NH}}_{\text{4}}{\text{Cl}}_{(aq)}{\text{ + NaOH}}_{(aq)}\to {\text{ H}}_2{\text{O}}_{(l)}+{\text{ NH}}_3{}_{(g)}+{\text{ NaCl}}_{(aq)}$

Before

$0.093\text{5 mol}$

$0.125\text{ mol}$

$0$

$0$ $0$

Change

$-0.093\text{5 mol}$

$-0.09\text{35 mol}$ $+0.09\text{35 mol}$

$+0.093\text{5 mol}$

$+0.0935\text{m}\text{o}\text{l}$

______________________________________________________________________________

After

$0$

$\text{0}\text{.11}6\text{m}\text{o}\text{l}$

$0.09\text{35 mol}$

$0.093\text{5 mol}$

$0.0\text{935 mol}$

Possibility II: if NaOH runs out first

Balanced equation

${\text{NH}}_{\text{4}}{\text{Cl}}_{(aq)}{\text{ + NaOH}}_{(aq)}\to {\text{ H}}_{\text{2}}{\text{O}}_{(l)}+{\text{ NH}}_3{}_{(g)}+{\text{ NaCl}}_{(aq)}$

Before

$0.09\text{35 mol}$

$0.125\text{ mol}$

$0$

$0$ $0$

Change

$-0.1\text{25 mol}$

$-0.1\text{25 mol}$ $+0.12\text{5 mol}$

$+0.1\text{25 mol}$

$+0.1\text{25 mol}$

______________________________________________________________________________

After

$-0.03\text{15 mol}$

$0$

$0.125\text{ mol}$

$0.12\text{5 mol}$

$0.12\text{5 mol}$

According to BCA tables, NaOH is not the limiting reactant as, to react with all the NaOH, we need $\text{0}\text{.0315 mol}$ 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 = $0.09\text{35 mol}$ × $\text{18}\text{.016 g/mol}$ = $\text{1}\text{.68 g}$

Mass of NH₃ produce = $0.093\text{5 mol}$ × $\text{17}\text{.034 g/mol}$ = $\text{1}\text{.59 g}$

Mass of NaCl produce = $0.09\text{35 mol}$ × $\text{58}\text{.44 g/mol}$ = $\text{5}\text{.46 g}$.

Interpretation Introduction

(d)

Interpretation:

Using Before-Change-After (BCA) tables, the limiting reactant should be determined in the given unbalanced equation, supposing that exactly $15.0\text{ g}$ 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

${\text{N}}_2 + 3{\text{H}}_2 \to 2{\text{NH}}_3$

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 = $\text{6}\text{.65 }\text{g}$

Mass of CO₂ produce = $\text{7}\text{.88 }\text{g}$.

Explanation of Solution

Number of moles of Fe₂ O₃ = $\frac{5.0\text{ g}}{\text{159}\text{.7 g/mol}}$ = $0.031\text{3 mol}$

Number of moles of CO = $\frac{5.0\text{ g}}{\text{28}\text{.01 g/mol}}$ = $0.179\text{ mol}$

Possibility I: if Fe₂ O₃ runs out first

Balanced equation

${\text{Fe}}_{\text{2}}{\text{O}}_3{}_{(s)}{\text{ + 3CO}}_{(g)}\to 2{\text{Fe}}_{(s)}{\text{ + 3CO}}_{2(g)}$

Before

$0.031\text{3 mol}$ $0.179\text{ mol}$

$0$

$0$

Change

$-0.031\text{3 mol}$ $-0.0939\text{ mol}$

$+0.62\text{6 mol}$

$+0.093\text{9 mol}$

____________________________________________________________________________

After

$0$ $\text{0}\text{.0851 mol}$

$0.62\text{6 mol}$

$0.62\text{6 mol}$

Possibility II: if CO runs out first

Balanced equation

${\text{Fe}}_{\text{2}}{\text{O}}_3{}_{(s)}{\text{ + 3CO}}_{(g)}\to 2{\text{Fe}}_{(s)}{\text{ + 3CO}}_{2(g)}$

Before

$0.0313\text{ mol}$ $0.179\text{ mol}$

$0$

$0$

Change

$-0.059\text{7 mol}$ $-0.179\text{ mol}$

$+0.11\text{9 mol}$

$+0.179\text{ mol}$

____________________________________________________________________________

After

$-0.028\text{4 mol}$

$0$

$0.11\text{9 mol}$

$0.179\text{ mol}$

According to BCA tables, CO is not the limiting reactant as, to react with all the CO, we need $\text{0}\text{.134 }\text{m}\text{o}\text{l}$ 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 = $0.11\text{9 mol}$ × $\text{55}\text{.85 g/mol}$ = $\text{6}\text{.65 g}$

Mass of CO₂ produce = $0.179\text{ mol}$ × $\text{44}\text{.01 g/mol}$ = $\text{7}\text{.88 g}$.