Chapter 8, Problem 96E

Interpretation Introduction

Interpretation:

The mass of ${\text{KO}}_2$ is to be determined, which can produce enough amount of oxygen for the users present in mine.

Concept introduction:

When mass of product is given and the mass of reactant is to be determined, then follow the following solution map:

$\boxed{\begin{array}{l}\text{given}\\ \text{mass}\text{of}\text{product}\end{array}}\to \boxed{\text{moles of product}}\underset{\begin{array}{l}\text{By}\text{using}\\ \text{conversion}\\ \text{factor}\end{array}}{\to }\boxed{\text{moles}\text{of}\text{reactant}}\to \boxed{\text{mass}\text{of}\text{reactant}}$

The conversion factor used to convert the moles of product into the moles of reactant is basically the stoichiometric relationship between moles of product and moles of reactant. The conversion factor is obtained from the balanced chemical equation. The mass of reactant or product is converted into their corresponding moles by the use of their corresponding molar masses.

In this given problem, the reactant is ${\text{KO}}_2$ and product is ${\text{O}}_{\text{2}}$. To calculate the mass of ${\text{KO}}_2$ required in grams for the production of given mass of ${\text{O}}_{\text{2}}$, follow the following solution map:

$\boxed{\text{mass}\text{of}{\text{O}}_{\text{2}}}\to \boxed{{\text{moles of O}}_{\text{2}}}\underset{\begin{array}{l}\text{By}\text{using}\\ \text{conversion}\\ \text{factor}\end{array}}{\to }\boxed{\text{moles}\text{of}{\text{KO}}_2}\to \boxed{\text{mass}\text{of}{\text{KO}}_{\text{2}}}$