Chapter 12, Problem 12.82P

Interpretation Introduction

Interpretation:

The mass of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ produced has to be calculated.

Concept Introduction:

The formula to evaluate moles from molarity is given as follows:

  $n=MV$

Here,

$M$ denotes molarity.

$n$ denotes number of moles.

$V$ denotes volume.

Mass percent is calculated by the formula as follows:

  $\text{Mass percent}=\frac{\text{Mass of solute}}{\text{Total Mass }}\left(100\text{ \%}\right)$

Answer to Problem 12.82P

The mass of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ produced is $2.7744\text{ g}$.

Explanation of Solution

The conversion factor to convert liters to milliliters is as follows:

  $\text{1 L}=\text{1000 mL}$

Hence convert $\text{55 mL}$ to $\text{L}$ as follows:

  $\begin{array}{c}\text{Volume}=\left(\text{55 mL}\right)\left(\frac{1\text{ L}}{1000\text{ mL}}\right)\\ =0.055\text{ L}\end{array}$

Similarly, convert $\text{100 mL}$ to $\text{L}$ as follows:

  $\begin{array}{c}\text{Volume}=\left(\text{100 mL}\right)\left(\frac{1\text{ L}}{1000\text{ mL}}\right)\\ =0.1\text{ L}\end{array}$

The formula to evaluate moles from molarity is given as follows:

  $n=MV$        (1)

Substitute $0.560\text{ mol/L}$ for $M$ and $0.055\text{ L}$ for $V$ in equation (1) to calculate moles of ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$.

  $\begin{array}{c}n=\left(0.560\text{ mol/L}\right)\left(0.055\text{ L}\right)\\ =0.0308\text{ mol}\end{array}$

Substitute $0.963\text{ mol/L}$ for $M$ and $0.1\text{ L}$ for $V$ in equation (1) to calculate moles of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$.

  $\begin{array}{c}n=\left(0.963\text{ mol/L}\right)\left(0.1\text{ L}\right)\\ =0.0963\text{ mol}\end{array}$

The balanced reaction is given as follows:

  $\left[\begin{array}{l}{\text{3CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\left(aq\right)+2{\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\left(aq\right)\\ +16{\text{H}}^{+}\left(aq\right)\end{array}\right]\to \left[\begin{array}{c}3{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\left(aq\right)+\\ 4{\text{Cr}}^{2+}\left(aq\right)+11{\text{H}}_2\text{O}\left(l\right)\end{array}\right]$        (2)

According to the stoichiometry of equation (2), ${\text{3 mol HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is produced by ${\text{2 mol Cr}}_{\text{2}}{\text{O}}_7^{2-}$, thus moles of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ produced by $0.0963\text{ mol}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}=\left(0.0308{\text{ mol Cr}}_{\text{2}}{\text{O}}_7^{2-}\right)\left(\frac{{\text{3 mol HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}}{{\text{2 mol Cr}}_{\text{2}}{\text{O}}_7^{2-}}\right)\\ =0.0462{\text{ mol HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\end{array}$

Similarly, moles of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ produced by $0.0963{\text{ mol CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$  is calculated as follows:

  $\begin{array}{c}{\text{Moles of HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}=\left(0.0963{\text{ mol CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\right)\left(\frac{{\text{3 mol HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}}{{\text{3 mol CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}}\right)\\ =0.0963{\text{ mol HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\end{array}$

Since ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$ yields less moles of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$, it acts as limiting reagent and thus moles of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ produced is $0.0462\text{ mol}$ .

The formula to convert mass in gram to moles is as follows:

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{Molar mass}\right)$        (3)

Substitute $0.0462\text{ mol}$ for moles and $\text{60}\text{.052 g/mol}$ for molar mass of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ in equation (3).

  $\begin{array}{c}\text{Mass}=\left(0.0462\text{ mol}\right)\left(\text{60}\text{.052 g/mol}\right)\\ =2.7744\text{ g}\end{array}$