Chapter 9, Problem 34A

(a)

Interpretation Introduction

Interpretation:The limiting reactant is to be determined. The mass of products produced is to be calculated.

Concept introduction: The limiting reactant is the reactant which limits the amount of product formed in the reaction.

Answer to Problem 34A

The limiting reactant is $\text{HCl}$ . The mass of ${\text{AlCl}}_{\text{3}}$ and ${\text{H}}_{\text{2}}$ produced is $19.635\text{g}$ and $0.414\text{g}$ .

Explanation of Solution

The given unbalanced chemical equation is,

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

The balanced chemical equation is written as,

  $\text{Al}\left(s\right)+3\text{HCl}\left(aq\right)\to {\text{AlCl}}_3\left(aq\right)+\frac{3}{2}{\text{H}}_2\left(g\right)$

The number of moles of $\text{Al}$ is calculated as,

  $n_{\text{Al}}=\frac{m_{\text{Al}}}{M_{\text{Al}}}$

Where,

• $m_{\text{Al}}$ is the given mass of $\text{Al}$ .
• $M_{\text{Al}}$ is the molar mass of $\text{Al}$$\left(26.98\text{g/mol}\right)$ .

Substitute the value of given mass and molar mass of $\text{Al}$ in the above formula.

  $\begin{array}{c}{n}_{\text{Al}}=\frac{15.0\text{g}}{26.98\text{g/mol}}\\ =0.556\text{mol}\end{array}$

The number of moles of $\text{HCl}$ is calculated as,

  $n_{\text{HCl}}=\frac{m_{\text{HCl}}}{M_{\text{HCl}}}$

Where,

• $m_{\text{HCl}}$ is the given mass of $\text{HCl}$ .
• $M_{\text{HCl}}$ is the molar mass of $\text{HCl}$$\left(36.46\text{g/mol}\right)$ .

Substitute the value of given mass and molar mass of $\text{HCl}$ in the above formula.

  $\begin{array}{c}{n}_{\text{HCl}}=\frac{15.0\text{g}}{36.46\text{g/mol}}\\ =0.411\text{mol}\end{array}$

The molar ratio of $\text{Al}$ and $\text{HCl}$ is $1:3$ . The number of moles of $\text{HCl}$ required to completely react with $0.556\text{mol}$ of $\text{Al}$ is $1.668\text{mol}$ but available number of moles of HCl is 0.411 mol therefore, $\text{HCl}$ is the limiting reactant.

The amount of ${\text{AlCl}}_{\text{3}}$ is calculated by the formula,

  $m_{{\text{AlCl}}_3}=\frac{1}{3}\times n_{\text{HCl}}\times M_{{\text{AgCl}}_{\text{3}}}$

Where,

• $M_{{\text{AgCl}}_{\text{3}}}$ is the molar mass of ${\text{AgCl}}_{\text{3}}\left(143.32\text{g/mol}\right)$ .

Substitute the number of moles of $\text{HCl}$ and molar mass of ${\text{AgCl}}_{\text{3}}$ in the above formula.

  $\begin{array}{c}{m}_{{\text{AlCl}}_3}=\frac{1}{3}\times 0.411\text{mol}\times 143.32\text{g/mol}\\ =19.635\text{g}\end{array}$

The amount of ${\text{H}}_{\text{2}}$ is calculated by the formula,

  $m_{{\text{H}}_{\text{2}}}=\frac{1}{2}\times n_{\text{HCl}}\times M_{{\text{H}}_{\text{2}}}$

Where,

• $M_{{\text{H}}_{\text{2}}}$ is the molar mass of ${\text{H}}_{\text{2}}\left(2.016\text{g/mol}\right)$ .

Substitute the number of moles of $\text{HCl}$ and molar mass of ${\text{H}}_{\text{2}}$ in the above formula.

  $\begin{array}{c}{m}_{{\text{H}}_{\text{2}}}=\frac{1}{2}\times 0.411\text{mol}\times 2.016\text{g/mol}\\ =0.414\text{g}\end{array}$

(b)

Interpretation Introduction

Interpretation: The limiting reactant is to be determined. The mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ and ${\text{H}}_{\text{2}}\text{O}$ produced is to be calculated.

Concept introduction: The limiting reactant is the reactant which limits the amount of product formed in the reaction.

Answer to Problem 34A

The limiting reactant is $\text{NaOH}$ . The mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ and ${\text{H}}_{\text{2}}\text{O}$ produced is $19.875\text{g}$ and $3.375\text{g}$ .

Explanation of Solution

The given unbalanced chemical equation is,

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

The balanced chemical equation is written as,

  $2\text{NaOH}\left(aq\right)+{\text{CO}}_2\left(g\right)\to {\text{Na}}_2{\text{CO}}_3\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

The number of moles of $\text{NaOH}$ is calculated as,

  $n_{\text{NaOH}}=\frac{m_{\text{NaOH}}}{M_{\text{NaOH}}}$

Where,

• $m_{\text{NaOH}}$ is the given mass of $\text{NaOH}$ .
• $M_{\text{NaOH}}$ is the molar mass of $\text{NaOH}$$\left(40.0\text{g/mol}\right)$ .

Substitute the value of given mass and molar mass of $\text{NaOH}$ in the above formula.

  $\begin{array}{c}{n}_{\text{NaOH}}=\frac{15.0\text{g}}{40.0\text{g/mol}}\\ =0.375\text{mol}\end{array}$

The number of moles of ${\text{CO}}_2$ is calculated as,

  $n_{{\text{CO}}_2}=\frac{m_{{\text{CO}}_2}}{M_{{\text{CO}}_2}}$

Where,

• $m_{{\text{CO}}_2}$ is the given mass of ${\text{CO}}_2$ .
• $M_{{\text{CO}}_2}$ is the molar mass of ${\text{CO}}_2$$\left(44.01\text{g/mol}\right)$ .

Substitute the value of given mass and molar mass of ${\text{CO}}_2$ in the above formula.

  $\begin{array}{c}{n}_{{\text{CO}}_2}=\frac{15.0\text{g}}{44.01\text{g/mol}}\\ =0.341\text{mol}\end{array}$

The molar ratio of $\text{NaOH}$ and ${\text{CO}}_2$ is $2:1$ . The number of moles of $\text{NaOH}$ required to completely react with $0.341\text{mol}$ of ${\text{CO}}_2$ is $0.685\text{mol}$ . Therefore, $\text{NaOH}$ is the limiting reactant.

The amount of ${\text{Na}}_2{\text{CO}}_{\text{3}}$ is calculated by the formula,

  $m_{{\text{Na}}_2{\text{CO}}_{\text{3}}}=\frac{1}{2}\times n_{\text{NaOH}}\times M_{{\text{Na}}_2{\text{CO}}_{\text{3}}}$

Where,

• $M_{{\text{Na}}_2{\text{CO}}_{\text{3}}}$ is the molar mass of ${\text{Na}}_2{\text{CO}}_{\text{3}}\left(106.0\text{g/mol}\right)$ .

Substitute the number of moles of $\text{NaOH}$ and molar mass of ${\text{Na}}_2{\text{CO}}_{\text{3}}$ in the above formula.

  $\begin{array}{c}{m}_{{\text{Na}}_2{\text{CO}}_{\text{3}}}=\frac{1}{2}\times 0.375\text{mol}\times 106.0\text{g/mol}\\ =19.875\text{g}\end{array}$

The amount of ${\text{H}}_2\text{O}$ is calculated by the formula,

  $m_{{\text{H}}_2\text{O}}=\frac{1}{2}\times n_{\text{NaOH}}\times M_{{\text{H}}_2\text{O}}$

Where,

• $M_{{\text{H}}_2\text{O}}$ is the molar mass of ${\text{H}}_2\text{O}\left(18.0\text{g/mol}\right)$ .

Substitute the number of moles of $\text{NaOH}$ and molar mass of ${\text{H}}_2\text{O}$ in the above formula.

  $\begin{array}{c}{m}_{{\text{H}}_2\text{O}}=\frac{1}{2}\times 0.375\text{mol}\times 18.0\text{g/mol}\\ =3.375\text{g}\end{array}$

(c)

Interpretation Introduction

Interpretation: The limiting reactant is to be determined. The mass of ${\text{PbCl}}_{\text{2}}$ and ${\text{HNO}}_{\text{3}}$ produced is to be calculated.

Concept introduction:

Answer to Problem 34A

The limiting reactant is $\text{Pb}{\left({\text{NO}}_3\right)}_2$ . The mass of ${\text{PbCl}}_{\text{2}}$ and ${\text{HNO}}_{\text{3}}$ produced is $12.6\text{g}$ and $5.71\text{g}$ .

Explanation of Solution

The given unbalanced chemical equation is,

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

The balanced chemical equation is written as,

  $\text{Pb}{\left({\text{NO}}_3\right)}_2\left(aq\right)+2\text{HCl}\left(aq\right)\to {\text{PbCl}}_2\left(s\right)+2{\text{HNO}}_3\left(aq\right)$

The number of moles of $\text{Pb}{\left({\text{NO}}_3\right)}_2$ is calculated as,

  $n_{\text{Pb}{\left({\text{NO}}_3\right)}_2}=\frac{m_{\text{Pb}{\left({\text{NO}}_3\right)}_2}}{M_{\text{Pb}{\left({\text{NO}}_3\right)}_2}}$

Where,

• $m_{\text{Pb}{\left({\text{NO}}_3\right)}_2}$ is the given mass of $\text{Pb}{\left({\text{NO}}_3\right)}_2$ .
• $M_{\text{Pb}{\left({\text{NO}}_3\right)}_2}$ is the molar mass of $\text{Pb}{\left({\text{NO}}_3\right)}_2$$\left(331.2\text{g/mol}\right)$ .

Substitute the value of given mass and molar mass of $\text{Pb}{\left({\text{NO}}_3\right)}_2$ in the above formula.

  $\begin{array}{c}{n}_{\text{Pb}{\left({\text{NO}}_3\right)}_2}=\frac{15.0\text{g}}{331.2\text{g/mol}}\\ =0.0453\text{mol}\end{array}$

The number of moles of $\text{HCl}$ is calculated as,

  $n_{\text{HCl}}=\frac{m_{\text{HCl}}}{M_{\text{HCl}}}$

Where,

• $m_{\text{HCl}}$ is the given mass of $\text{HCl}$ .
• $M_{\text{HCl}}$ is the molar mass of $\text{HCl}$$\left(36.46\text{g/mol}\right)$ .

Substitute the value of given mass and molar mass of $\text{Al}$ in the above formula.

  $\begin{array}{c}{n}_{\text{HCl}}=\frac{15.0\text{g}}{36.46\text{g/mol}}\\ =0.411\text{mol}\end{array}$

The molar ratio of $\text{Pb}{\left({\text{NO}}_3\right)}_2$ and $\text{HCl}$ is $1:2$ . The number of moles of $\text{HCl}$ required to completely react with $0.0453\text{mol}$ of $\text{Pb}{\left({\text{NO}}_3\right)}_2$ is $0.0906\text{mol}$ . Therefore, $\text{Pb}{\left({\text{NO}}_3\right)}_2$ is the limiting reactant.

The amount of ${\text{PbCl}}_2$ is calculated by the formula,

  $m_{{\text{PbCl}}_2}=n_{\text{Pb}{\left({\text{NO}}_3\right)}_2}\times M_{{\text{PbCl}}_2}$

Where,

• $M_{{\text{PbCl}}_2}$ is the molar mass of ${\text{PbCl}}_2\left(278.1\text{g/mol}\right)$ .

Substitute the number of moles of $\text{Pb}{\left({\text{NO}}_3\right)}_2$ and molar mass of ${\text{PbCl}}_2$ in the above formula.

  $\begin{array}{c}{m}_{{\text{PbCl}}_2}=0.0453\text{mol}\times 278.1\text{g/mol}\\ =12.6\text{g}\end{array}$

The amount of ${\text{HNO}}_{\text{3}}$ is calculated by the formula,

  $m_{{\text{HNO}}_{\text{3}}}=2\times n_{\text{Pb}{\left({\text{NO}}_3\right)}_2}\times M_{{\text{HNO}}_{\text{3}}}$

Where,

• $M_{{\text{HNO}}_{\text{3}}}$ is the molar mass of ${\text{HNO}}_{\text{3}}\left(63.01\text{g/mol}\right)$ .

Substitute the number of moles of $\text{Pb}{\left({\text{NO}}_3\right)}_2$ and molar mass of ${\text{HNO}}_{\text{3}}$ in the above formula.

  $\begin{array}{c}{m}_{{\text{HNO}}_{\text{3}}}=2\times 0.0453\text{mol}\times 63.01\text{g/mol}\\ =5.71\text{g}\end{array}$

(d)

Interpretation Introduction

Interpretation: The limiting reactant is to be determined. The mass of KI produced is to be calculated.

Concept introduction:

Answer to Problem 34A

The limiting reactant is ${\text{I}}_{\text{2}}$ . The mass of product produced is $19.62\text{g}$ .

Explanation of Solution

The given unbalanced chemical equation is,

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

The balanced chemical equation is written as,

  $2\text{K}\left(s\right)+{\text{I}}_2\left(s\right)\to 2\text{KI}\left(s\right)$

The number of moles of $\text{K}$ is calculated as,

  $n_{\text{K}}=\frac{m_{\text{K}}}{M_{\text{K}}}$

Where,

• $m_{\text{K}}$ is the given mass of $\text{K}$ .
• $M_{\text{K}}$ is the molar mass of $\text{K}$$\left(39.1\text{g/mol}\right)$ .

Substitute the value of given mass and molar mass of $\text{K}$ in the above formula.

  $\begin{array}{c}{n}_{\text{K}}=\frac{15.0\text{g}}{39.1\text{g/mol}}\\ =0.384\text{mol}\end{array}$

The number of moles of ${\text{I}}_2$ is calculated as,

  $n_{{\text{I}}_2}=\frac{m_{{\text{I}}_2}}{M_{{\text{I}}_2}}$

Where,

• $m_{{\text{I}}_2}$ is the given mass of ${\text{I}}_2$ .
• $M_{{\text{I}}_2}$ is the molar mass of ${\text{I}}_2$$\left(253.81\text{g/mol}\right)$ .

Substitute the value of given mass and molar mass of $\text{Al}$ in the above formula.

  $\begin{array}{c}{n}_{{\text{I}}_2}=\frac{15.0\text{g}}{253.81\text{g/mol}}\\ =0.0591\text{mol}\end{array}$

The molar ratio of $\text{K}$ and ${\text{I}}_2$ is $2:1$ . The number of moles of $\text{K}$ required to completely react with $0.0591\text{mol}$ of ${\text{I}}_2$ is $0.1182\text{mol}$ . Therefore, ${\text{I}}_2$ is the limiting reactant.

The amount of $\text{KI}$ is calculated by the formula,

  $m_{\text{KI}}=2\times n_{{\text{I}}_2}\times M_{\text{KI}}$

Where,

• $M_{\text{KI}}$ is the molar mass of $\text{KI}\left(166.0\text{g/mol}\right)$ .

Substitute the number of moles of ${\text{I}}_2$ and molar mass of $\text{KI}$ in the above formula.

  $\begin{array}{c}{m}_{\text{KI}}=2\times 0.0591\text{mol}\times 166.0\text{g/mol}\\ =19.62\text{g}\end{array}$