Chapter 11, Problem 119A

Interpretation Introduction

Interpretation:

The reactant present in excess and the mass of the reactant needs to be determined.

Concept introduction:

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

Answer to Problem 119A

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

Hydroquinone is excess reactant. 45.1 mg of hydroquinone was left.

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. And hydroquinone acts as excess reactant. Only 0.49 mmoles of benzoquinone was formed. So, according to reaction stoichiometry, 0.49 mmoles of benzoquinone consume 0.49 mmoles of hydroquinone.

The excess hydroquinone = total mmoles of hydroquinone − mmoles of hydroquinone consumed

= 0.9- 0.49

= 0.41 mmoles

Mass of hydroquinone left $=0.41\times 110=45.1mg$.