Chapter 14, Problem 37PE

(a)

Interpretation Introduction

Interpretation:

Moles of ${\text{H}}_2\text{O}$ produced from $15.0\text{ mL}$ of $0.250M\text{ HCl}$ have to be determined.

Concept Introduction:

Stoichiometry describes quantitative relationships between reactants and products in any chemical reactions. In these problems, amount of one species is determined with help of known amount of another species.

Explanation of Solution

Given reaction occurs as follows:

  $2{\text{KMnO}}_4\left(aq\right)+16\text{HCl}\left(aq\right)\to 2{\text{MnCl}}_2\left(aq\right)+5{\text{Cl}}_2\left(g\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)+2\text{KCl}\left(aq\right)$

Expression for molarity of $\text{HCl}$ is as follows:

  $\text{Molarity of HCl}=\frac{\text{Moles of HCl}}{\text{Volume}\left(\text{L}\right)\text{ of HCl}}$        (1)

Rearrange equation (1) for moles of $\text{HCl}$.

  $\text{Moles of HCl}=\left(\text{Molarity of HCl}\right)\left(\text{Volume of HCl}\right)$        (2)

Substitute $15.0\text{ mL}$ for volume and $0.250M$ for molarity of $\text{HCl}$ in equation (2).

  $\begin{array}{c}\text{Moles of HCl}=\left(0.250M\right)\left(15.0\text{ mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.00375\text{ mol}\end{array}$

According to balanced chemical equation, two moles of ${\text{KMnO}}_4$ react with sixteen moles of $\text{HCl}$ and produce two moles of ${\text{MnCl}}_2$, eight moles of ${\text{H}}_2\text{O}$ and two moles of $\text{KCl}$. Moles of ${\text{H}}_2\text{O}$ produced from 0.00375 moles of $\text{HCl}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of H}}_2\text{O}=\left(\frac{{\text{8 mol H}}_2\text{O}}{\text{16 mol HCl}}\right)\left(\text{0}\text{.00375 mol HCl}\right)\\ =0.001875\text{ mol}\end{array}$

Hence, amount of ${\text{H}}_2\text{O}$ is $0.001875\text{ mol}$.

(b)

Interpretation Introduction

Interpretation:

Volume of $0.150M{\text{ KMnO}}_4$ produced from $1.85{\text{ mol MnCl}}_2$ have to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction occurs as follows:

  $2{\text{KMnO}}_4\left(aq\right)+16\text{HCl}\left(aq\right)\to 2{\text{MnCl}}_2\left(aq\right)+5{\text{Cl}}_2\left(g\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)+2\text{KCl}\left(aq\right)$

According to balanced chemical equation, two moles of ${\text{KMnO}}_4$ react with sixteen moles of $\text{HCl}$ and produce two moles of ${\text{MnCl}}_2$, eight moles of ${\text{H}}_2\text{O}$ and two moles of $\text{KCl}$. Amount of ${\text{KMnO}}_4$ produced by 1.85 moles of ${\text{MnCl}}_2$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of KMnO}}_4=\left(\frac{2{\text{ mol KMnO}}_4}{2{\text{ mol MnCl}}_2}\right)\left(1.85{\text{ mol MnCl}}_2\right)\\ =1.85\text{ mol}\end{array}$

Expression for molarity of ${\text{KMnO}}_4$ is as follows:

  ${\text{Molarity of KMnO}}_4=\frac{{\text{Moles of KMnO}}_4}{\text{Volume}\left(\text{L}\right){\text{ of KMnO}}_4}$        (3)

Rearrange equation (3) for volume of ${\text{KMnO}}_4$.

  ${\text{Volume of KMnO}}_4=\frac{{\text{Moles of KMnO}}_4}{{\text{Molarity of KMnO}}_4}$        (4)

Substitute $1.85\text{ mol}$ for moles and $0.150M$ for molarity of ${\text{KMnO}}_4$ in equation (4).

  $\begin{array}{c}{\text{Volume of KMnO}}_4=\frac{1.85\text{ mol}}{0.150M}\\ =12.33\text{ L}\end{array}$

Hence, volume of ${\text{KMnO}}_4$ is $12.33\text{ L}$.

(c)

Interpretation Introduction

Interpretation:

Volume of $\text{2}\text{.50 }M\text{ HCl}$ required to produce $125\text{ mL}$ of $\text{0}\text{.525 }M\text{ KCl}$ have to be determined.

Concept Introduction:

Expression for molarity equation is as follows:

  $M_1{V}_1=M_2{V}_2$        (5)

Here,

$M_1$ is molarity of one solution.

$V_1$ is volume of one solution.

$M_2$ is molarity of another solution.

$V_2$ is volume of another solution.

Explanation of Solution

Rearrange equation (5) for $V_2$.

  $V_2=\frac{M_1{V}_1}{M_2}$        (6)

Substitute $\text{0}\text{.525 }M$ for $M_1$, $125\text{ mL}$ for $V_1$ and $\text{2}\text{.50 }M$ for $M_2$ in equation (6) for volume of $\text{HCl}$.

  $\begin{array}{c}\text{Volume of HCl}=\frac{\left(0.525M\right)\left(\text{125 mL}\right)}{\text{2}\text{.50 }M}\\ =26.25\text{ mL}\end{array}$

Hence volume of $\text{HCl}$ required is $26.25\text{ mL}$.

(d)

Interpretation Introduction

Interpretation:

Molarity of $\text{HCl}$ solution if its $22.20\text{ mL}$ reacts with $15.60\text{ mL}$ of $\text{0}\text{.250 }M{\text{ KMnO}}_4$ has to be determined.

Concept Introduction:

Refer to part (c).

Explanation of Solution

Rearrange equation (5) for $M_2$.

  $M_2=\frac{M_1{V}_1}{V_2}$        (7)

Substitute $\text{0}\text{.250 }M$ for $M_1$, $15.60\text{ mL}$ for $V_1$ and $22.20\text{ mL}$ for $V_2$ in equation (7) for molarity of $\text{HCl}$.

  $\begin{array}{c}\text{Molarity of HCl}=\frac{\left(\text{0}\text{.250 }M\right)\left(\text{15}\text{.60 mL}\right)}{\text{22}\text{.20 mL}}\\ =0.176M\end{array}$

Hence molarity of $\text{HCl}$ required is $0.176M$.

(e)

Interpretation Introduction

Interpretation:

Volume $\left(\text{L}\right)$ of ${\text{Cl}}_{\text{2}}$ gas at STP produced by reaction of $125\text{ mL}$ of $\text{2}\text{.5 }M\text{ HCl}$ has to be determined.

Concept Introduction:

Expression for ideal gas equation is as follows:

  $PV=nRT$        (8)

Here,

$P$ is pressure of gas.

$V$ is volume of gas.

$n$ is moles of gas.

$R$ is universal gas constant.

$T$ is absolute temperature of gas.

Explanation of Solution

Given reaction occurs as follows:

  $2{\text{KMnO}}_4\left(aq\right)+16\text{HCl}\left(aq\right)\to 2{\text{MnCl}}_2\left(aq\right)+5{\text{Cl}}_2\left(g\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)+2\text{KCl}\left(aq\right)$

Expression for molarity of $\text{HCl}$ is as follows:

  $\text{Molarity of HCl}=\frac{\text{Moles of HCl}}{\text{Volume}\left(\text{L}\right)\text{ of HCl solution}}$        (9)

Rearrange equation (9) for moles of $\text{HCl}$.

  $\text{Moles of HCl}=\left[\begin{array}{c}\left(\text{Molarity of HCl}\right)\\ \left(\text{Volume of HCl solution}\right)\end{array}\right]$        (10)

Substitute $2.5M$ for molarity and $125\text{ mL}$ for volume of $\text{HCl}$ solution in equation (10).

  $\begin{array}{c}\text{Moles of HCl}=\left(2.5M\right)\left(\text{125 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.3125\text{ mol}\end{array}$

Since 16 moles of $\text{HCl}$ produce 5 moles of ${\text{Cl}}_{\text{2}}$, amount of ${\text{Cl}}_2$ produced by 0.3125 moles of $\text{HCl}$ is calculated as follows:

  $\begin{array}{c}{\text{Amount of Cl}}_2=\left(\frac{{\text{5 mol Cl}}_2}{\text{16 mol HCl}}\right)\left(0.3125\text{ mol HCl}\right)\\ =0.0976\text{ mol}\end{array}$

Rearrange equation (8) for $V$.

  $V=\frac{nRT}{P}$        (11)

Substitute $273\text{ K}$ for $T$, $0.0976\text{ mol}$ for $n$, $\text{0}\text{.082057 L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$ and $1\text{ atm}$ for $P$ in equation (11) for volume of ${\text{Cl}}_2$.

  $\begin{array}{c}{\text{Volume of Cl}}_2=\frac{\left(0.0976\text{ mol}\right)\left(\text{0}\text{.082057 L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(273\text{ K}\right)}{1\text{ atm}}\\ =2.186\text{ L}\end{array}$

Hence, volume of ${\text{Cl}}_2$ is $2.186\text{ L}$.

(f)

Interpretation Introduction

Interpretation:

Volume $\left(\text{L}\right)$ of ${\text{Cl}}_{\text{2}}$ gas at STP produced by reaction of $15.0\text{ mL}$ of $\text{0}\text{.750 }M\text{ HCl}$ and $12.0\text{ mL}$ of $0.550M{\text{ KMnO}}_4$ has to be determined.

Concept Introduction:

Refer to part (e).

Explanation of Solution

Given reaction occurs as follows:

  $2{\text{KMnO}}_4\left(aq\right)+16\text{HCl}\left(aq\right)\to 2{\text{MnCl}}_2\left(aq\right)+5{\text{Cl}}_2\left(g\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)+2\text{KCl}\left(aq\right)$

Expression for molarity of $\text{HCl}$ is as follows:

  $\text{Molarity of HCl}=\frac{\text{Moles of HCl}}{\text{Volume}\left(\text{L}\right)\text{ of HCl solution}}$        (9)

Rearrange equation (9) for moles of $\text{HCl}$.

  $\text{Moles of HCl}=\left[\begin{array}{c}\left(\text{Molarity of HCl}\right)\\ \left(\text{Volume of HCl solution}\right)\end{array}\right]$        (10)

Substitute $\text{0}\text{.750 }M$ for molarity and $15.0\text{ mL}$ for volume of $\text{HCl}$ solution in equation (10).

  $\begin{array}{c}\text{Moles of HCl}=\left(\text{0}\text{.750 }M\right)\left(\text{15}\text{.0 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.01125\text{ mol}\end{array}$

Expression for molarity of ${\text{KMnO}}_4$ is as follows:

  ${\text{Molarity of KMnO}}_4=\frac{{\text{Moles of KMnO}}_4}{\text{Volume}\left(\text{L}\right){\text{ of KMnO}}_4}$        (3)

Rearrange equation (3) for moles of ${\text{KMnO}}_4$.

  ${\text{Moles of KMnO}}_4=\left[\begin{array}{c}\left({\text{Molarity of KMnO}}_4\right)\\ \left({\text{Volume of KMnO}}_4\right)\end{array}\right]$        (12)

Substitute $12.0\text{ mL}$ for molarity and $0.550M$ for molarity of ${\text{KMnO}}_4$ in equation (12).

  $\begin{array}{c}{\text{Moles of KMnO}}_4=\left(0.550M\right)\left(12.0\text{ mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.0066\text{ mol}\end{array}$

Since 16 moles of $\text{HCl}$ form 5 moles of ${\text{Cl}}_{\text{2}}$, amount of ${\text{Cl}}_{\text{2}}$ produced by 0.01125 moles of $\text{HCl}$ is calculated as follows:

  $\begin{array}{c}{\text{Amount of Cl}}_{\text{2}}=\left(\frac{{\text{5 mol Cl}}_{\text{2}}}{16\text{ mol HCl}}\right)\left(0.01125\text{ mol HCl}\right)\\ =0.003516\text{ mol}\end{array}$

Since 2 moles of ${\text{KMnO}}_4$ form 5 moles of ${\text{Cl}}_{\text{2}}$, amount of ${\text{Cl}}_{\text{2}}$ produced by 0.0066 moles of ${\text{KMnO}}_4$ is calculated as follows:

  $\begin{array}{c}{\text{Amount of Cl}}_{\text{2}}=\left(\frac{{\text{5 mol Cl}}_{\text{2}}}{{\text{2 mol KMnO}}_4}\right)\left(0.0066{\text{ mol KMnO}}_4\right)\\ =0.0165\text{ mol}\end{array}$

Since $\text{HCl}$ produces less amount of ${\text{Cl}}_{\text{2}}$ than ${\text{KMnO}}_{\text{4}}$, $\text{HCl}$ acts as limiting reactant and amount of ${\text{Cl}}_{\text{2}}$ will be produced in accordance with its amount only.

Rearrange equation (8) for $V$.

  $V=\frac{nRT}{P}$        (11)

Substitute $273\text{ K}$ for $T$, $0.003516\text{ mol}$ for $n$, $\text{0}\text{.082057 L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$ and $1\text{ atm}$ for $P$ in equation (11) for volume of ${\text{Cl}}_2$.

  $\begin{array}{c}{\text{Volume of Cl}}_2=\frac{\left(0.003516\text{ mol}\right)\left(\text{0}\text{.082057 L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(273\text{ K}\right)}{1\text{ atm}}\\ =0.079\text{ L}\end{array}$

Hence, volume of ${\text{Cl}}_2$ is $0.079\text{ L}$.