Chapter 11, Problem 120A

Interpretation Introduction

Interpretation:

The milligrams of Benzoquinone produced needs to be determined.

Concept introduction:

Limiting reactant is the reactant which does not remain after the completion of reaction.

Excess reactant is the reactant which is present in excess amount and left over after completion of reaction.

Answer to Problem 120A

52.92 mg of benzoquinone will be produced.

Explanation of Solution

Molecular mass of hydroquinone is 110 g/mol.

$\begin{array}{c}\text{number}\text{of}\text{moles}\text{of}\text{Hydroquinone=}\frac{\text{Mass}\text{of}\text{Hydroquinone}}{\text{Molar}\text{mass}\text{of}\text{Hydroquinone}}\\ \text{=}\frac{\text{100}\text{mg}}{\text{110}\text{g/mol}}\\ \text{=}\text{0}\text{.90}\text{mmoles}\end{array}$

Or,

$\begin{array}{c}\text{number}\text{of}\text{moles}\text{of}\text{H2O2=}\frac{\text{Mass}\text{of}\text{H2O2}}{\text{Molar}\text{mass}\text{of}\text{H2O2}}\\ \text{=}\frac{\text{50}\text{mg}}{\text{34}\text{g/mol}}\\ \text{=}\text{1}\text{.47}\text{mmoles}\end{array}$

The chemical reaction is as follows:

From the stoichiometry of reaction.

For hydroquinone,

$\begin{array}{c}\text{1}\text{moles}\text{of}\text{Hydroquinone}\text{produce}\text{=}\text{1}\text{moles}\text{of}\text{benzoquinone}\\ \text{0}\text{.90}\text{mmoles}\text{of}\text{Hydroquinone}\text{produce}\text{=}\text{1×0}\text{.9}\text{mmoles}\text{of}\text{benzoquinone}\\ \text{0}\text{.90}\text{mmoles}\text{of}\text{Hydroquinone}\text{produce}\text{=}\text{0}\text{.9}\text{mmoles}\text{of}\text{benzoquinone}\end{array}$

For H₂O_2.

$\begin{array}{c}\text{3}\text{moles}\text{of}\text{H2O2}\text{produce}\text{=}\text{1}\text{moles}\text{of}\text{Benzoquinone}\\ \text{1}\text{moles}\text{of}\text{H2O2}\text{produce}\text{=}\frac{\text{1}}{\text{3}}\text{moles}\text{of}\text{Benzoquinone}\\ \text{1}\text{.47×1}\text{mmoles}\text{of}\text{H2O2}\text{produce}\text{=}\text{1}\text{.47×}\frac{\text{1}}{\text{3}}\text{mmoles}\text{of}\text{Benzoquinone}\\ \text{1}\text{.47}\text{mmoles}\text{of}\text{H2O2}\text{produce}\text{=}\text{0}\text{.49}\text{mmoles}\text{of}\text{Benzoquinone}\end{array}$

The reactant which produces least amount of product is limiting reagent. So, $H2O2$ act as the limiting reagent. Hence, 0.49 mmoles of benzoquinone will be produced.

Mass of benzoquinone = moles X molar mass

= $0.49\times 108=52.9\text{2 mg of benzoquinone}$