Chapter 17, Problem 116A

Interpretation Introduction

Interpretation: The amount of oxygen formed by the decomposition of hydrogen peroxide is to be found.

Concept introduction: According to Avogadro’s law, 1 mole of every substance weighs equal to molecular mass and contains Avogadro’s number $6.023\times {10}^{23}$ of particles.

The number of moles of any substance is found by dividing the given amount of substance by the molecular mass of the substance.

Answer to Problem 116A

The amount of oxygen formed on the decomposition of $25.0\text{ g}$ of hydrogen peroxide is $\text{11}\text{.7g}$

Explanation of Solution

The given chemical reaction is:

  ${\text{2H}}_{\text{2}}{\text{O}}_{\text{2}}\text{(l)}\to {\text{2H}}_{\text{2}}{\text{O(l)+O}}_{\text{2}}\text{(g)}$

According to stoichiometry, it is given that for every $\text{2}$ moles of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ , $\text{1}$ mole of ${\text{O}}_{\text{2}}$ is formed.

given mass of hydrogen peroxide = $25.0\text{ g}$

Molar mass of hydrogen peroxide = $\text{34}\text{.01 g/mol}$

Putting in the values we get:

  ${\text{moles of hydrogen peroxide (H}}_2{\text{O}}_2\text{) = }\frac{25.0\text{ g}}{\text{34}\text{.01 g/mol}}=0.735\text{mol}$

  $\text{2}$ moles of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ leads to formation of = $\text{1}$ mole of ${\text{O}}_{\text{2}}$

Thus, $0.735\text{ mol}$ of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ will lead to formation of = $\frac{\text{1 mol}}{\text{2 mol}}\times 0.735\text{ mol= 0}\text{.367 mol}$

Amount of ${\text{O}}_{\text{2}}$ formed = $\text{0}\text{.367 mol}\times \text{ 32 g/mol = 11}\text{.7g}$