Chapter 11, Problem 118A

Interpretation Introduction

Interpretation:

The given chemical equation needs to be balanced. Also, if 100 mg of hydroquinone and 50 mg of H₂O₂ react, the limiting reactant needs to be determined.

Concept introduction:

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

Answer to Problem 118A

H₂O₂ is limiting reactant.

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.