Chapter 14, Problem 38PE

(a)

Interpretation Introduction

Interpretation:

Moles of ${\text{H}}_2\text{O}$ produced from $25.0\text{ mL}$ of $0.150M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$ 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:

  ${\text{K}}_2{\text{CO}}_3\left(aq\right)+2{\text{HC}}_2{\text{H}}_3{\text{O}}_2\left(aq\right)\to 2{\text{KC}}_2{\text{H}}_3{\text{O}}_2\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$

Expression for molarity of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ is as follows:

  ${\text{Molarity of HC}}_2{\text{H}}_3{\text{O}}_2=\frac{{\text{Moles of HC}}_2{\text{H}}_3{\text{O}}_2}{\text{Volume}\left(\text{L}\right){\text{ of HC}}_2{\text{H}}_3{\text{O}}_2\text{ solution}}$        (1)

Rearrange equation (1) for moles of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$.

  ${\text{Moles of HC}}_2{\text{H}}_3{\text{O}}_2=\left[\begin{array}{c}\left({\text{Molarity of HC}}_2{\text{H}}_3{\text{O}}_2\right)\\ \left({\text{Volume of HC}}_2{\text{H}}_3{\text{O}}_2\text{ solution}\right)\end{array}\right]$        (2)

Substitute $25.0\text{ mL}$ for volume and $0.150M$ for molarity of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ in equation (2).

  $\begin{array}{c}{\text{Moles of HC}}_2{\text{H}}_3{\text{O}}_2=\left(0.150M\right)\left(25.0\text{ mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.00375\text{ mol}\end{array}$

According to balanced chemical equation, one mole of ${\text{K}}_2{\text{CO}}_3$ reacts with two moles of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ and produce two moles of ${\text{KC}}_2{\text{H}}_3{\text{O}}_2$, one mole of ${\text{H}}_{\text{2}}\text{O}$ and one mole of ${\text{CO}}_{\text{2}}$. Moles of ${\text{H}}_2\text{O}$ produced from 0.00375 moles of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of H}}_2\text{O}=\left(\frac{{\text{1 mol H}}_2\text{O}}{{\text{2 mol HC}}_2{\text{H}}_3{\text{O}}_2}\right)\left(0.00375{\text{ mol HC}}_2{\text{H}}_3{\text{O}}_2\right)\\ =0.001875\text{ mol}\end{array}$

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

(b)

Interpretation Introduction

Interpretation:

Volume of $0.210M{\text{ K}}_2{\text{CO}}_3$ produced from $17.5{\text{ mol KC}}_2{\text{H}}_3{\text{O}}_2$ have to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction occurs as follows:

  ${\text{K}}_2{\text{CO}}_3\left(aq\right)+2{\text{HC}}_2{\text{H}}_3{\text{O}}_2\left(aq\right)\to 2{\text{KC}}_2{\text{H}}_3{\text{O}}_2\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$

According to balanced chemical equation, one mole of ${\text{K}}_2{\text{CO}}_3$ reacts with two moles of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ and produce two moles of ${\text{KC}}_2{\text{H}}_3{\text{O}}_2$, one mole of ${\text{H}}_{\text{2}}\text{O}$ and one mole of ${\text{CO}}_{\text{2}}$. Amount of ${\text{K}}_2{\text{CO}}_3$ produced by 17.5 moles of ${\text{KC}}_2{\text{H}}_3{\text{O}}_2$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of K}}_2{\text{CO}}_3=\left(\frac{1{\text{ mol K}}_2{\text{CO}}_3}{2{\text{ mol KC}}_2{\text{H}}_3{\text{O}}_2}\right)\left(17.5{\text{ mol KC}}_2{\text{H}}_3{\text{O}}_2\right)\\ =8.75\text{ mol}\end{array}$

Expression for molarity of ${\text{K}}_2{\text{CO}}_3$ is as follows:

  ${\text{Molarity of K}}_2{\text{CO}}_3=\frac{{\text{Moles of K}}_2{\text{CO}}_3}{\text{Volume}\left(\text{L}\right){\text{ of K}}_2{\text{CO}}_3\text{ solution}}$        (3)

Rearrange equation (3) for volume of ${\text{K}}_2{\text{CO}}_3$.

  ${\text{Volume of K}}_2{\text{CO}}_3=\frac{{\text{Moles of K}}_2{\text{CO}}_3}{{\text{Molarity of K}}_2{\text{CO}}_3}$        (4)

Substitute $8.75\text{ mol}$ for moles and $0.210M$ for molarity of ${\text{K}}_2{\text{CO}}_3$ in equation (4).

  $\begin{array}{c}{\text{Volume of K}}_2{\text{CO}}_3=\frac{8.75\text{ mol}}{0.210M}\\ =41.67\text{ L}\end{array}$

Hence, volume of ${\text{K}}_2{\text{CO}}_3$ is $41.67\text{ L}$.

(c)

Interpretation Introduction

Interpretation:

Volume of $\text{1}\text{.25 }M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$ required to react with $\text{75}\text{.2 mL}$ of $\text{0}\text{.750 }M{\text{ K}}_2{\text{CO}}_3$ 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 $\text{0}\text{.750 }M$ for $M_1$, $\text{75}\text{.2 mL}$ for $V_1$ and $\text{1}\text{.25 }M$ for $M_2$ in equation (6) for volume of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$.

  $\begin{array}{c}{\text{Volume of HC}}_2{\text{H}}_3{\text{O}}_2=\frac{\left(\text{0}\text{.750 }M\right)\left(\text{75}\text{.2 mL}\right)}{\text{1}\text{.25 }M}\\ =45.12\text{ mL}\end{array}$

Hence, volume of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ required is $45.12\text{ mL}$.

(d)

Interpretation Introduction

Interpretation:

Molarity of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ solution if its $10.15\text{ mL}$ reacts with $\text{18}\text{.50 mL}$ of $\text{0}\text{.250 }M{\text{ K}}_2{\text{CO}}_3$ 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 $M_1$, $\text{18}\text{.50 mL}$ for $V_1$ and $10.15\text{ mL}$ for $V_2$ in equation (7) for molarity of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$.

  $\begin{array}{c}{\text{Molarity of HC}}_2{\text{H}}_3{\text{O}}_2=\frac{\left(\text{0}\text{.250 }M\right)\left(\text{18}\text{.50 mL}\right)}{10.15\text{ mL}}\\ =0.456M\end{array}$

Hence, molarity of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ required is $0.456M$.

(e)

Interpretation Introduction

Interpretation:

Volume $\left(\text{L}\right)$ of ${\text{CO}}_{\text{2}}$ gas at STP produced by reaction of $105\text{ mL}$ of $\text{1}\text{.5 }M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$ 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:

  ${\text{K}}_2{\text{CO}}_3\left(aq\right)+2{\text{HC}}_2{\text{H}}_3{\text{O}}_2\left(aq\right)\to 2{\text{KC}}_2{\text{H}}_3{\text{O}}_2\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$

Expression for molarity of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ is as follows:

  ${\text{Molarity of HC}}_2{\text{H}}_3{\text{O}}_2=\frac{{\text{Moles of HC}}_2{\text{H}}_3{\text{O}}_2}{\text{Volume}\left(\text{L}\right){\text{ of HC}}_2{\text{H}}_3{\text{O}}_2\text{ solution}}$        (9)

Rearrange equation (9) for moles of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$.

  ${\text{Moles of HC}}_2{\text{H}}_3{\text{O}}_2=\left[\begin{array}{c}\left({\text{Molarity of HC}}_2{\text{H}}_3{\text{O}}_2\right)\\ \left({\text{Volume of HC}}_2{\text{H}}_3{\text{O}}_2\text{ solution}\right)\end{array}\right]$        (10)

Substitute $1.5M$ for molarity and $105\text{ mL}$ for volume of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ solution in equation (10).

  $\begin{array}{c}{\text{Moles of HC}}_2{\text{H}}_3{\text{O}}_2=\left(1.5M\right)\left(\text{105 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.1575\text{ mol}\end{array}$

Since 2 moles of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ produce 1 moleof ${\text{CO}}_{\text{2}}$, amount of ${\text{CO}}_{\text{2}}$ produced by 0.1575 moles of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ is calculated as follows:

  $\begin{array}{c}{\text{Amount of CO}}_2=\left(\frac{{\text{1 mol CO}}_2}{{\text{2 mol HC}}_2{\text{H}}_3{\text{O}}_2}\right)\left(0.1575{\text{ mol HC}}_2{\text{H}}_3{\text{O}}_2\right)\\ =0.07875\text{ mol}\end{array}$

Rearrange equation (8) for $V$.

  $V=\frac{nRT}{P}$        (11)

Substitute $273\text{ K}$ for $T$, $0.07875\text{ mol}$ 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 ${\text{CO}}_2$.

  $\begin{array}{c}{\text{Volume of CO}}_2=\frac{\left(0.07875\text{ mol}\right)\left(\text{0}\text{.082057 L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(273\text{ K}\right)}{1\text{ atm}}\\ =1.775\text{ L}\end{array}$

Hence, volume of ${\text{CO}}_2$ is $1.775\text{ L}$.

(f)

Interpretation Introduction

Interpretation:

Volume $\left(\text{L}\right)$ of ${\text{CO}}_2$ gas at STP produced by reaction of $25.0\text{ mL}$ of $\text{0}\text{.350 }M{\text{ K}}_2{\text{CO}}_3$ and $25.0\text{ mL}$ of $0.250M{\text{ HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ has to be determined.

Concept Introduction:

Refer to part (e).

Explanation of Solution

Given reaction occurs as follows:

  ${\text{K}}_2{\text{CO}}_3\left(aq\right)+2{\text{HC}}_2{\text{H}}_3{\text{O}}_2\left(aq\right)\to 2{\text{KC}}_2{\text{H}}_3{\text{O}}_2\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$

Expression for molarity of ${\text{K}}_2{\text{CO}}_3$ is as follows:

  ${\text{Molarity of K}}_2{\text{CO}}_3=\frac{{\text{Moles of K}}_2{\text{CO}}_3}{\text{Volume}\left(\text{L}\right){\text{ of K}}_2{\text{CO}}_3\text{ solution}}$        (3)

Rearrange equation (9) for moles of ${\text{K}}_2{\text{CO}}_3$.

  ${\text{Moles of K}}_2{\text{CO}}_3=\left[\begin{array}{c}\left({\text{Molarity of K}}_2{\text{CO}}_3\right)\\ \left({\text{Volume of K}}_2{\text{CO}}_3\text{ solution}\right)\end{array}\right]$        (12)

Substitute $\text{0}\text{.350 }M$ for molarity and $25.0\text{ mL}$ for volume of ${\text{K}}_2{\text{CO}}_3$ solution in equation (10).

  $\begin{array}{c}{\text{Moles of K}}_2{\text{CO}}_3=\left(\text{0}\text{.350 }M\right)\left(\text{25}\text{.0 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.00875\text{ mol}\end{array}$

Expression for molarity of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ is as follows:

  ${\text{Molarity of HC}}_2{\text{H}}_3{\text{O}}_2=\frac{{\text{Moles of HC}}_2{\text{H}}_3{\text{O}}_2}{\text{Volume}\left(\text{L}\right){\text{ of HC}}_2{\text{H}}_3{\text{O}}_2\text{ solution}}$        (9)

Rearrange equation (9) for moles of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$.

  ${\text{Moles of HC}}_2{\text{H}}_3{\text{O}}_2=\left[\begin{array}{c}\left({\text{Molarity of HC}}_2{\text{H}}_3{\text{O}}_2\right)\\ \left({\text{Volume of HC}}_2{\text{H}}_3{\text{O}}_2\text{ solution}\right)\end{array}\right]$        (10)

Substitute $\text{0}\text{.250 }M$ for molarity and $25.0\text{ mL}$ for volume of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ solution in equation (10).

  $\begin{array}{c}{\text{Moles of HC}}_2{\text{H}}_3{\text{O}}_2=\left(\text{0}\text{.250 }M\right)\left(\text{25}\text{.0 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.00625\text{ mol}\end{array}$

Since 1 mole of ${\text{K}}_2{\text{CO}}_3$ forms1 moleof ${\text{CO}}_2$, amount of ${\text{CO}}_2$ produced by 0.00875 moles of ${\text{K}}_2{\text{CO}}_3$ is calculated as follows:

  $\begin{array}{c}{\text{Amount of CO}}_{\text{2}}=\left(\frac{{\text{1 mol CO}}_{\text{2}}}{1{\text{ mol K}}_2{\text{CO}}_3}\right)\left(0.00875{\text{ mol K}}_2{\text{CO}}_3\right)\\ =0.00875\text{ mol}\end{array}$

Since 2 moles of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ form 1 moleof ${\text{CO}}_2$, amount of ${\text{CO}}_2$ produced by 0.0066 moles of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ is calculated as follows:

  $\begin{array}{c}{\text{Amount of CO}}_{\text{2}}=\left(\frac{{\text{1 mol CO}}_{\text{2}}}{{\text{2 mol HC}}_2{\text{H}}_3{\text{O}}_2}\right)\left(0.00625{\text{ mol HC}}_2{\text{H}}_3{\text{O}}_2\right)\\ =0.003125\text{ mol}\end{array}$

Since ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ produces less amount of ${\text{CO}}_2$ than ${\text{K}}_2{\text{CO}}_3$, ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ acts as limiting reactant and amount of ${\text{CO}}_2$ will be produced in accordance with its amount only.

Rearrange equation (8) for $V$.

  $V=\frac{nRT}{P}$        (11)

Substitute $273\text{ K}$ for $T$, $0.003125\text{ mol}$ 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 ${\text{CO}}_2$.

  $\begin{array}{c}{\text{Volume of CO}}_2=\frac{\left(0.003125\text{ mol}\right)\left(\text{0}\text{.082057 L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(273\text{ K}\right)}{1\text{ atm}}\\ =0.070\text{ L}\end{array}$

Hence, volume of ${\text{CO}}_2$ is $0.070\text{ L}$.