Chapter 4, Problem 22E

The reaction of potassium superoxide, ${\text{KO}}_{\text{2}}\text{,}$ is used in life-support systems to replace ${\text{CO}}_{\text{2}}\left(\text{g}\right)$ in expired air with ${\text{O}}_{\text{2}}\text{g}$ . The unbalanced chemical equation for the reaction is given below.
${\text{KO}}_{\text{2}}\left(\text{s}\right){\text{ + CO}}_{\text{2}}\left(\text{g}\right){\text{ + K}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right){\text{ + O}}_{\text{2}}\left(\text{g}\right)$
a. How many moles of ${\text{O}}_{\text{2}}\text{g}$ are produced by the reaction of 88.0 g ${\text{CO}}_{\text{2}}\left(\text{g}\right)$ with excess ${\text{KO}}_{\text{2}}$ (s)?
b. How many grams of ${\text{KO}}_{\text{2}}$ (s) are consumed per $\text{1}{\text{.000 × 10}}^{\text{2}}{\text{g CO}}_{\text{2}}\left(\text{g}\right)$ removed from expired air?
c. How many moles of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{2}}$ are produced per milligram of ${\text{KO}}_{\text{2}}$ consumed?