Chapter 14, Problem 36PE

(a)

Interpretation Introduction

Interpretation:

Moles of ${\text{Na}}_2{\text{SO}}_4$ produced from $3.6{\text{ mol H}}_2{\text{SO}}_4$ 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:

  $2\text{NaOH}\left(aq\right)+{\text{H}}_2{\text{SO}}_4\left(aq\right)\to {\text{Na}}_2{\text{SO}}_4\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

According to balanced chemical equation, two moles of $\text{NaOH}$ react with one mole of ${\text{H}}_2{\text{SO}}_4$ and produce one mole of ${\text{Na}}_2{\text{SO}}_4$ and two moles of ${\text{H}}_2\text{O}$. Since stoichiometric ratio between ${\text{H}}_2{\text{SO}}_4$ and ${\text{Na}}_2{\text{SO}}_4$ is $1:1$, amount of ${\text{Na}}_2{\text{SO}}_4$ produced from 3.6 moles of ${\text{H}}_2{\text{SO}}_4$ is calculated as follows:

  $\begin{array}{c}{\text{Amount of Na}}_2{\text{SO}}_4=\left(\frac{1{\text{ mol Na}}_2{\text{SO}}_4}{{\text{1 mol H}}_2{\text{SO}}_4}\right)\left(\text{3}{\text{.6 mol H}}_2{\text{SO}}_4\right)\\ =3.6\text{ mol}\end{array}$

Hence, amount of ${\text{Na}}_2{\text{SO}}_4$ is $3.6\text{ mol}$.

(b)

Interpretation Introduction

Interpretation:

Moles of ${\text{H}}_2\text{O}$ produced from $0.025\text{ mol NaOH}$ have to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction occurs as follows:

  $2\text{NaOH}\left(aq\right)+{\text{H}}_2{\text{SO}}_4\left(aq\right)\to {\text{Na}}_2{\text{SO}}_4\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

According to balanced chemical equation, two moles of $\text{NaOH}$ react with one mole of ${\text{H}}_2{\text{SO}}_4$ and produce one mole of ${\text{Na}}_2{\text{SO}}_4$ and two moles of ${\text{H}}_2\text{O}$. Since stoichiometric ratio between $\text{NaOH}$ and ${\text{H}}_2\text{O}$ is $2:2$, amount of ${\text{H}}_2\text{O}$ produced from 0.025 moles of $\text{NaOH}$ is calculated as follows:

  $\begin{array}{c}{\text{Amount of H}}_2\text{O}=\left(\frac{2{\text{ mol H}}_2\text{O}}{\text{2 mol NaOH}}\right)\left(\text{0}\text{.025 mol NaOH}\right)\\ =\text{0}\text{.025 mol}\end{array}$

Hence, amount of ${\text{H}}_2\text{O}$ is $\text{0}\text{.025 mol}$.

(c)

Interpretation Introduction

Interpretation:

Moles of $\text{NaOH}$ required to react with $2.50\text{ L}$ of $\text{0}\text{.125 }M{\text{ H}}_2{\text{SO}}_4$ have to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction occurs as follows:

  $2\text{NaOH}\left(aq\right)+{\text{H}}_2{\text{SO}}_4\left(aq\right)\to {\text{Na}}_2{\text{SO}}_4\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Expression to calculate molarity of ${\text{H}}_2{\text{SO}}_4$ is as follows:

  ${\text{Molarity of H}}_2{\text{SO}}_4=\frac{{\text{Moles of H}}_2{\text{SO}}_4}{\text{Volume }\left(\text{L}\right){\text{ of H}}_2{\text{SO}}_4\text{ solution}}$        (1)

Rearrange equation (1) for moles of ${\text{H}}_2{\text{SO}}_4$.

  ${\text{Moles of H}}_2{\text{SO}}_4=\left[\begin{array}{c}\left({\text{Molarity of H}}_2{\text{SO}}_4\right)\\ \left({\text{Volume of H}}_2{\text{SO}}_4\text{ solution}\right)\end{array}\right]$        (2)

Substitute $\text{0}\text{.125 }M$ for molarity and $2.50\text{ L}$ for volume of ${\text{H}}_2{\text{SO}}_4$ solution in equation (2).

  $\begin{array}{c}{\text{Moles of H}}_2{\text{SO}}_4=\left(\text{0}\text{.125 }M\right)\left(\text{2}\text{.50 L}\right)\\ =0.3125\text{ mol}\end{array}$

According to balanced chemical equation, two moles of $\text{NaOH}$ react with one mole of ${\text{H}}_2{\text{SO}}_4$ and produce one mole of ${\text{Na}}_2{\text{SO}}_4$ and two moles of ${\text{H}}_2\text{O}$. Since stoichiometric ratio between ${\text{H}}_2{\text{SO}}_4$ and $\text{NaOH}$ is $1:2$, amount of $\text{NaOH}$ required to react with 0.3125 moles of ${\text{H}}_2{\text{SO}}_4$ is calculated as follows:

  $\begin{array}{c}\text{Amount of NaOH}=\left(\frac{\text{2 mol NaOH}}{{\text{1 mol H}}_2{\text{SO}}_4}\right)\left(\text{0}{\text{.3125 mol H}}_2{\text{SO}}_4\right)\\ =0.625\text{ mol}\end{array}$

Hence, amount of $\text{NaOH}$ required is $0.625\text{ mol}$.

(d)

Interpretation Introduction

Interpretation:

Grams of ${\text{Na}}_2{\text{SO}}_4$ obtained from $\text{25 mL}$ of $\text{0}\text{.050 }M\text{ NaOH}$ have to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction occurs as follows:

  $2\text{NaOH}\left(aq\right)+{\text{H}}_2{\text{SO}}_4\left(aq\right)\to {\text{Na}}_2{\text{SO}}_4\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Expression to calculate molarity of $\text{NaOH}$ is as follows:

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

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

  $\text{Moles of NaOH}=\left[\begin{array}{c}\left(\text{Molarity of NaOH}\right)\\ \left(\text{Volume of NaOH solution}\right)\end{array}\right]$        (4)

Substitute $\text{0}\text{.050 }M$ for molarity and $\text{25 mL}$ for volume of $\text{NaOH}$ solution in equation (4).

  $\begin{array}{c}\text{Moles of NaOH}=\left(\text{0}\text{.050 }M\right)\left(\text{25 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.00125\text{ mol}\end{array}$

According to balanced chemical equation, two moles of $\text{NaOH}$ react with one mole of ${\text{H}}_2{\text{SO}}_4$ and produce one mole of ${\text{Na}}_2{\text{SO}}_4$ and two moles of ${\text{H}}_2\text{O}$. Since stoichiometric ratio between ${\text{Na}}_2{\text{SO}}_4$ and $\text{NaOH}$ is $1:2$, amount of ${\text{Na}}_2{\text{SO}}_4$ required to react with 0.0625 moles of $\text{NaOH}$ is calculated as follows:

  $\begin{array}{c}{\text{Amount of Na}}_2{\text{SO}}_4=\left(\frac{1{\text{ mol Na}}_2{\text{SO}}_4}{\text{2 mol NaOH}}\right)\left(\text{0}\text{.00125 mol NaOH}\right)\\ =0.000625\text{ mol}\end{array}$

Expression for mass of ${\text{Na}}_2{\text{SO}}_4$ is as follows:

  ${\text{Mass of Na}}_2{\text{SO}}_4=\left[\begin{array}{c}\left({\text{Moles of Na}}_2{\text{SO}}_4\right)\\ \left({\text{Molar mass of Na}}_2{\text{SO}}_4\right)\end{array}\right]$        (5)

Substitute $0.000625\text{ mol}$ for moles and $\text{142}\text{.04 g/mol}$ for molar mass of ${\text{Na}}_2{\text{SO}}_4$ in equation (5).

  $\begin{array}{c}{\text{Mass of Na}}_2{\text{SO}}_4=\left(0.000625\text{ mol}\right)\left(142.04\text{ g/mol}\right)\\ =0.088775\text{ g}\end{array}$

Hence, $0.088775\text{ g}$ of ${\text{Na}}_2{\text{SO}}_4$ is required.

(e)

Interpretation Introduction

Interpretation:

Volume of $\text{0}\text{.250 }M{\text{ H}}_2{\text{SO}}_4$ required to react with $\text{25}\text{.5 mL}$ of $\text{0}\text{.750 }M\text{ NaOH}$ have to be determined.

Concept Introduction:

Expression for molarity equation is as follows:

  $M_1{V}_1=M_2{V}_2$        (6)

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 (6) for $V_2$.

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

Substitute $\text{0}\text{.750 }M$ for $M_1$, $\text{25}\text{.5 mL}$ for $V_1$ and $\text{0}\text{.250 }M$ for $M_2$ in equation (7) for volume of ${\text{H}}_2{\text{SO}}_4$.

  $\begin{array}{c}{\text{Volume of H}}_2{\text{SO}}_4=\frac{\left(0.750M\right)\left(\text{25}\text{.5 mL}\right)}{\text{0}\text{.250 }M}\\ =76.5\text{ mL}\end{array}$

Hence volume of ${\text{H}}_2{\text{SO}}_4$ required is $76.5\text{ mL}$.

(f)

Interpretation Introduction

Interpretation:

Molarity of $\text{NaOH}$ formed by reaction of $\text{48}\text{.20 mL}$ of $\text{NaOH}$ with $\text{35}\text{.72 mL}$ of $\text{0}\text{.125 }M{\text{ H}}_2{\text{SO}}_4$ have to be determined.

Concept Introduction:

Refer to part (e).

Explanation of Solution

Rearrange equation (6) for $M_2$.

  $M_2=\frac{M_1{V}_1}{V_2}$        (8)

Substitute $\text{0}\text{.125 }M$ for $M_1$, $\text{48}\text{.20 mL}$ for $V_1$ and $\text{35}\text{.72 mL}$ for $V_2$ in equation (8) for molarity of $\text{NaOH}$.

  $\begin{array}{c}\text{Molarity of NaOH}=\frac{\left(\text{0}\text{.125 }M\right)\left(\text{48}\text{.20 mL}\right)}{\text{35}\text{.72 mL}}\\ =0.169M\end{array}$

Hence molarity of $\text{NaOH}$ required is $0.169M$.