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;  $\text{Al}\left(s\right)+\text{HCl}\left(aq\right)\to {\text{AlCl}}_3\left(aq\right)+{\text{H}}_2\left(g\right)$

msp;  $\text{NaOH}\left(aq\right)+{\text{CO}}_2\left(g\right)\to {\text{Na}}_2{\text{CO}}_3\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

msp;  $\text{Pb}{\left({\text{NO}}_3\right)}_2\left(aq\right)+\text{HCl}\left(aq\right)\to {\text{PbCl}}_2\left(s\right)+{\text{NHO}}_3\left(aq\right)$

msp;  $\text{K}\left(s\right)+{\text{I}}_2\left(s\right)\to \text{KI}\left(s\right)$

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

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 50QAP

The limiting reagent is HCl.

Mass of AlCl₃ produce = $18.3\text{g}$

Mass of H₂ produce = $0.415\text{g}$.

Explanation of Solution

Number of moles of Al = $\frac{15.0\text{ g}}{26.98\text{ g/mol}}$ = $0.556\text{m}\text{o}\text{l}$

Number of moles of HCl = $\frac{15.0\text{ g}}{36.458\text{ g/mol}}$ = $0.411\text{ mol}$

Possibility I: if Al runs out first

Balanced equation $2\text{A}{\text{l}}_{(s)}+6\text{H}\text{C}{\text{l}}_{(aq)}\to 2\text{A}\text{l}\text{C}{\text{l}}_{3(aq)}+3{\text{H}}_{2(g)}$

Before $0.556\text{m}\text{o}\text{l}$ $0.411\text{ mol}$ $0$ $0$

Change $-0.556\text{m}\text{o}\text{l}$ $-1.67\text{m}\text{o}\text{l}$ $+0.556\text{m}\text{o}\text{l}$ $+0.834\text{m}\text{o}\text{l}$

_________________________________________________________________________

After $0$ $-1.26\text{m}\text{o}\text{l}$ $0.556\text{m}\text{o}\text{l}$ $0.834\text{m}\text{o}\text{l}$

Possibility II: if HCl runs out first

Balanced equation$2\text{A}{\text{l}}_{(s)}+6\text{H}\text{C}{\text{l}}_{(aq)}\to 2\text{A}\text{l}\text{C}{\text{l}}_{3(aq)}+3{\text{H}}_{2(g)}$

Before$0.556\text{m}\text{o}\text{l}$ $0.411\text{ mol}$ $0$ $0$

Change$-0.137\text{m}\text{o}\text{l}$ $-0.411\text{ mol}$ $+0.137\text{m}\text{o}\text{l}$ $+0.206\text{m}\text{o}\text{l}$

_________________________________________________________________________

After $0.419\text{m}\text{o}\text{l}$ $0$ $0.137\text{m}\text{o}\text{l}$ $0.206\text{m}\text{o}\text{l}$

According to BCA tables, Al is not the limiting reactant as to react with all the Al we need $1.26\text{m}\text{o}\text{l}$ 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 = $0.137\text{m}\text{o}\text{l}$ × $133.33\text{g}/\text{m}\text{o}\text{l}$ = $18.3\text{g}$

Mass of H₂ produce = $0.206\text{m}\text{o}\text{l}$ × $2.016\text{g}/\text{m}\text{o}\text{l}$ = $0.415\text{g}$.

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

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 50QAP

The limiting reagent is NaOH

Mass of Na₂ CO₃ produce = $\text{19}\text{.9 }\text{g}$

Mass of H₂ O produce = $\text{3}\text{.39 }\text{g}$.

Explanation of Solution

Number of moles of NaOH = $\frac{15.0\text{ g}}{\text{39}\text{.998 g/mol}}$ = $0.375\text{m}\text{o}\text{l}$

Number of moles of CO₂ = $\frac{15.0\text{ g}}{\text{44}\text{.01 g/mol}}$ = $0.341\text{ mol}$

Possibility I: if NaOH runs out first

Balanced equation$2\text{N}\text{a}\text{O}{\text{H}}_{(aq)}{\text{ + CO}}_{2(g)}\to \text{N}{\text{a}}_2\text{C}{\text{O}}_{3(aq)}+{\text{H}}_2{\text{O}}_{(l)}$

Before$0.375\text{m}\text{o}\text{l}$ $0.341\text{ mol}$ $0$ $0$

Change$-0.375\text{m}\text{o}\text{l}$ $-0.188\text{m}\text{o}\text{l}$ $+0.188\text{m}\text{o}\text{l}$ $+0.188\text{m}\text{o}\text{l}$

______________________________________________________________________________

After $0$ $\text{0}\text{.153 }\text{m}\text{o}\text{l}$ $0.188\text{m}\text{o}\text{l}$ $0.188\text{m}\text{o}\text{l}$

Possibility II: if CO₂ runs out first

Balanced equation$2\text{N}\text{a}\text{O}{\text{H}}_{(aq)}{\text{ + CO}}_{2(g)}\to \text{N}{\text{a}}_2\text{C}{\text{O}}_{3(aq)}+{\text{H}}_2{\text{O}}_{(l)}$

Befor$0.375\text{m}\text{o}\text{l}$ $0.341\text{ mol}$ $0$ $0$

Change$-0.682\text{m}\text{o}\text{l}$ $-0.341\text{ mol}$ $+0.341\text{ mol}$ $+0.341\text{ mol}$

_________________________________________________________________________

After $-0.307\text{m}\text{o}\text{l}$ $0$ $0.341\text{ mol}$ $0.341\text{ mol}$

According to BCA tables, CO₂ is not the limiting reactant as to react with all the NaOH, we need $\text{0}\text{.307 }\text{m}\text{o}\text{l}$ 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 = $0.188\text{m}\text{o}\text{l}$ × $105.99\text{g}/\text{m}\text{o}\text{l}$ = $\text{19}\text{.9 }\text{g}$

Mass of H₂ O produce = $0.188\text{m}\text{o}\text{l}$ × $\text{18}\text{.016 }\text{g}/\text{m}\text{o}\text{l}$ = $\text{3}\text{.39 }\text{g}$.

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

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 50QAP

The limiting reagent is Pb(NO₃ )₂ Mass of PbCl₂ produce = $\text{12}\text{.6 }\text{g}$

Mass of HNO₃ produce = $\text{5}\text{.71 }\text{g}$.

Explanation of Solution

Number of moles of Pb(NO₃ )₂ = $\frac{15.0\text{ g}}{\text{331}\text{.22 g/mol}}$ = $0.0453\text{m}\text{o}\text{l}$

Number of moles of HCl = $\frac{15.0\text{ g}}{36.458\text{ g/mol}}$ = $0.411\text{ mol}$

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

Balanced equation $\text{P}\text{b}{(\text{N}{\text{O}}_3)}_2{}_{(aq)}{\text{ + 2HCl}}_{(aq)}\to {\text{ PbCl}}_2{}_{(s)}+{\text{ 2HNO}}_3{}_{(aq)}$

Before $0.0453\text{m}\text{o}\text{l}$ $0.411\text{ mol}$ $0$ $0$

Change $-0.0453\text{m}\text{o}\text{l}$ $-0.0906\text{m}\text{o}\text{l}$ $+0.0453\text{m}\text{o}\text{l}$ $+0.0906\text{m}\text{o}\text{l}$

______________________________________________________________________________

After $0$ $\text{0}\text{.320 }\text{m}\text{o}\text{l}$ $0.0453\text{m}\text{o}\text{l}$ $0.0906\text{m}\text{o}\text{l}$

Possibility II: if HCl runs out first

Balanced equation $\text{P}\text{b}{(\text{N}{\text{O}}_3)}_2{}_{(aq)}{\text{ + 2HCl}}_{(aq)}\to {\text{ PbCl}}_2{}_{(s)}+{\text{ 2HNO}}_3{}_{(aq)}$

Before $0.0453\text{m}\text{o}\text{l}$ $0.411\text{ mol}$ $0$ $0$

Change $-0.206\text{m}\text{o}\text{l}$ $-0.411\text{ mol}$ $+0.206\text{m}\text{o}\text{l}$ $+0.411\text{ mol}$

_________________________________________________________________________

After $-0.161\text{m}\text{o}\text{l}$ $0$ $0.206\text{ mol}$ $0.411\text{ mol}$

According to BCA tables, HCl is not the limiting reactant as, to react with all the HCl, we need $\text{0}\text{.161 }\text{m}\text{o}\text{l}$ 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 = $0.0453\text{m}\text{o}\text{l}$ × $\text{278}\text{.1 }\text{g}/\text{m}\text{o}\text{l}$ = $\text{12}\text{.6 }\text{g}$

Mass of HNO₃ produce = $0.0906\text{m}\text{o}\text{l}$ × $\text{63}\text{.018 }\text{g}/\text{m}\text{o}\text{l}$ = $\text{5}\text{.71 }\text{g}$.

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

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 50QAP

The limiting reagent is I₂.

Mass of KI produce = $\text{19}\text{.6 }\text{g}$.

Explanation of Solution

Number of moles of K = $\frac{15.0\text{ g}}{\text{39}\text{.00 g/mol}}$ = $0.385\text{m}\text{o}\text{l}$

Number of moles of I₂ = $\frac{15.0\text{ g}}{253.8\text{ g/mol}}$ = $0.0591\text{ mol}$

Possibility I: if K runs out first

Balanced equation $2{\text{K}}_{(s)}{\text{ + I}}_2{}_{(s)}\to 2\text{K}{\text{I}}_{(s)}$

Before $0.385\text{m}\text{o}\text{l}$ $0.0591\text{ mol}$ $0$

Change $-0.385\text{m}\text{o}\text{l}$ $-0.193\text{m}\text{o}\text{l}$ $+0.385\text{m}\text{o}\text{l}$

______________________________________________________

After $0$ $\text{-0}\text{.134 }\text{m}\text{o}\text{l}$ $0.385\text{m}\text{o}\text{l}$

Possibility II: if I₂ runs out first

Balanced equation $2{\text{K}}_{(s)}{\text{ + I}}_2{}_{(s)}\to 2\text{K}{\text{I}}_{(s)}$

Before $0.385\text{m}\text{o}\text{l}$ $0.0591\text{ mol}$ $0$

Change $-0.118\text{m}\text{o}\text{l}$ $-0.0591\text{ mol}$ $+0.118\text{m}\text{o}\text{l}$

______________________________________________________

After $0.267\text{m}\text{o}\text{l}$ $0$ $0.118\text{m}\text{o}\text{l}$

According to BCA tables, K is not the limiting reactant as, to react with all the K, we need $\text{0}\text{.134 }\text{m}\text{o}\text{l}$ 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 = $0.118\text{m}\text{o}\text{l}$ × $\text{165}\text{.9 }\text{g}/\text{m}\text{o}\text{l}$ = $\text{19}\text{.6 }\text{g}$.