Chapter 3, Problem 119E

Hydrogen peroxide is used as a cleansing agent in the treatment of cuts and abrasions for several reasons. It is an oxidizing agent that can directly kill many microorganisms; it decomposes on contact with blood, releasing elemental oxygen gas (which inhibits the growth of anaerobic microorganisms); and it foams on contact with blood, which provides a cleansing action. In the laboratory, small quantities of hydrogen peroxide can be prepared by the action of an acid on an alkaline earth metal peroxide, such as barium peroxide:

${\text{BaO}}_2\left(s\right)+2\text{HCl}\left(aq\right)\to {\text{H}}_{\text{2}}{\text{O}}_2\left(aq\right)+{\text{ BaCl}}_{\text{2}}\left(aq\right)$

What mass of hydrogen peroxide should result when 1.50 g barium peroxide is treated with 25.0 mL hydrochloric acid solution containing 0.0272 g HCI per mL? What mass of which reagent is left unreacted?

Interpretation Introduction

Interpretation: The amount of hydrogen peroxide that can be produced by the given chemical reaction is to be calculated. The mass of the unreacted reagent is to be calculated.

Concept introduction: The mass of a substance can be obtained by using the number of moles of the substance present and its molar mass. The formula used to calculate the mass of a given substance is,

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

To determine: The mass of hydrogen peroxide that can be produced by the given chemical reaction and the mass of the unreacted reagent.

Answer to Problem 119E

The amount of hydrogen peroxide that can be produced is $\underset{\_}{0.301g}$. The mass of the unreacted reagent, that is $\text{HCl}$, is $\underset{\_}{3.6\times 10^{-2}g}$.

Explanation of Solution

To determine: The mass of hydrogen peroxide that can be produced by the given chemical reaction

Given

The stated chemical reaction is,

${\text{BaO}}_2\left(s\right)+2\text{HCl}\left(aq\right)\to {\text{H}}_2{\text{O}}_2\left(aq\right)+{\text{BaCl}}_2\left(aq\right)$

The given mass of ${\text{BaO}}_2$ is $1.50\text{g}$.

The given volume of $\text{HCl}$ is $25.0\text{ml}$.

The solution contains $0.0272\text{g}\text{HCl}\text{per}\text{ml}$.

The molar mass of ${\text{BaO}}_2$ $\begin{array}{l}=\text{Ba}+2\text{O}\\ =\left(137+\left(2\times 16\right)\right)\text{g}/\text{mol}\\ =169\text{g}/\text{mol}\end{array}$

Formula

The number of moles of a substance is calculated by the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}\text{of}\text{the}\text{substance}}$

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

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{BaO}}_2=\frac{1.50\text{g}}{\text{169}\text{g}/\text{mol}}\\ =8.87\times {10}^{-3}\text{mol}\end{array}$

The solution contains $0.0272\text{g}\text{HCl}\text{per}\text{ml}$.

Therefore, the mass of $\text{HCl}$ $\begin{array}{l}=\left(0.0272\times 25.0\right)\text{g}\\ =0.68\text{g}\end{array}$

The molar mass of $\text{HCl}$ $\begin{array}{l}=\text{H}+\text{Cl}\\ =\left(1+35.5\right)\text{g}/\text{mol}\\ =36.5\text{g}/\text{mol}\end{array}$

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{HCl}=\frac{0.68\text{g}}{36.5\text{g}/\text{mol}}\\ =1.86\times {10}^{-2}\text{mol}\end{array}$

According to the stated reaction,

$1\text{mol}$ of ${\text{BaO}}_2$ reacts with $2\text{mol}$ of $\text{HCl}$.

$8.87\times {10}^{-3}\text{mol}$ of ${\text{BaO}}_2$ reacts with $\text{HCl}$ $\begin{array}{l}=\left(2\times 8.87\times {10}^{-3}\right)\text{mol}\\ =1.77\times {10}^{-2}\text{mol}\end{array}$

The moles of $\text{HCl}$ given are $1.86\times {10}^{-2}\text{mol}$.

Therefore, some amount of $\text{HCl}$ is left unreacted. It is the excess reactant.

Barium oxide $\left({\text{BaO}}_{\text{2}}\right)$ is the limiting reagent.

According to the stated reaction,

$1\text{mol}$ of ${\text{BaO}}_2$ produces $1\text{mol}$ of ${\text{H}}_2{\text{O}}_2$.

$8.87\times {10}^{-3}\text{mol}$ of ${\text{BaO}}_2$ produces ${\text{H}}_2{\text{O}}_2$ $\begin{array}{l}=\left(1\times 8.87\times {10}^{-3}\right)\text{mol}\\ =8.87\times {10}^{-3}\text{mol}\end{array}$

Molar mass of ${\text{H}}_2{\text{O}}_2$ $\begin{array}{l}=\text{2O}+2\text{H}\\ =\left(\left(2\times 16\right)+\left(2\times 1\right)\right)\text{g}/\text{mol}\\ =34\text{g}/\text{mol}\end{array}$

The mass of a substance is calculated by the formula,

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

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

$\begin{array}{c}\text{Mass}\text{of}{\text{H}}_2{\text{O}}_2=\left(\text{8}\text{.87}\times {\text{10}}^{-3}\text{mol}\right)\times \left(34\text{g}/\text{mol}\right)\\ =\underset{\_}{0.301g}\end{array}$

To determine: The unreacted amount of the starting material.

Given

The stated chemical reaction is,

${\text{BaO}}_2\left(s\right)+2\text{HCl}\left(aq\right)\to {\text{H}}_2{\text{O}}_2\left(aq\right)+{\text{BaCl}}_2\left(aq\right)$

The given mass of ${\text{BaO}}_2$ is $1.50\text{g}$.

The given volume of $\text{HCl}$ is $25.0\text{ml}$.

The solution contains $0.0272\text{g}\text{HCl}\text{per}\text{ml}$.

The molar mass of ${\text{BaO}}_2$ $\begin{array}{l}=\text{Ba}+2\text{O}\\ =\left(137+\left(2\times 16\right)\right)\text{g}/\text{mol}\\ =169\text{g}/\text{mol}\end{array}$

Formula

The number of moles of a substance is calculated by the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}\text{of}\text{the}\text{substance}}$

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

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{BaO}}_2=\frac{1.50\text{g}}{\text{169}\text{g}/\text{mol}}\\ =8.87\times {10}^{-3}\text{mol}\end{array}$

According to the stated reaction,

$1\text{mol}$ of ${\text{BaO}}_2$ reacts with $2\text{mol}$ of $\text{HCl}$.

$8.87\times {10}^{-3}\text{mol}$ of ${\text{BaO}}_2$ reacts with $\text{HCl}$ $\begin{array}{l}=\left(2\times 8.87\times {10}^{-3}\right)\text{mol}\\ =1.77\times {10}^{-2}\text{mol}\end{array}$

The moles of $\text{HCl}$ given are $1.86\times {10}^{-2}\text{mol}$.

Therefore, some amount of $\text{HCl}$ is left unreacted. It is the excess reactant.

The moles of $\text{HCl}$ left unreacted are calculated by the formula,

$\text{Unreacted}\text{moles}\text{of}\text{HCl}=\text{Given}\text{moles}\text{of}\text{HCl}-\text{Reacted}\text{moles}\text{of}\text{HCl}$

Substitute the values of the given moles of $\text{HCl}$ and the reacted moles of $\text{HCl}$ in the above expression.

$\begin{array}{c}\text{Unreacted}\text{moles}\text{of}\text{HCl}=\left(\text{1}\text{.86}\times {\text{10}}^{-2}-\text{1}\text{.77}\times {\text{10}}^{-2}\right)\text{mol}\\ =0.09\times {10}^{-2}\text{mol}\end{array}$

Molar mass of $\text{HCl}$ $\begin{array}{l}=\text{H}+\text{Cl}\\ =\left(1+35.5\right)\text{g}/\text{mol}\\ =36.5\text{g}/\text{mol}\end{array}$

The mass of a substance is calculated by the formula,

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

Substitute the value of the number of moles of $\text{HCl}$ and the molar mass of $\text{HCl}$ in the above expression.

$\begin{array}{c}\text{Mass}\text{of}\text{HCl}=\left(\text{0}\text{.09}\times {\text{10}}^{-2}\text{mol}\right)\times \left(36.5\text{g}/\text{mol}\right)\\ =\underset{\_}{3.3\times 10^{-2}g}\end{array}$

Conclusion

The amount of hydrogen peroxide that can be produced is $\underset{\_}{0.301g}$.

The unreacted reagent is $\text{HCl}$ and the unreacted mass of $\text{HCl}$ is $\underset{\_}{3.6\times 10^{-2}g}$.