The reaction between potassium superoxide, \(\mathrm{KO}_{2}\), and \(\mathrm{CO}_{2}\), $$ 4 \mathrm{KO}_{2}+2 \mathrm{CO}_{2} \longrightarrow 2 \mathrm{~K}_{2} \mathrm{CO}_{3}+3 \mathrm{O}_{2} $$ is used as a source of \(\mathrm{O}_{2}\) and absorber of \(\mathrm{CO}_{2}\) in self-contained breathing equipment used by rescue workers. (a) How many moles of \(\mathrm{O}_{2}\) are produced when \(0.400 \mathrm{~mol}\) of \(\mathrm{KO}_{2}\) reacts in this fashion? (b) How many grams of \(\mathrm{KO}_{2}\) are needed to form \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\) ? (c) How many grams of \(\mathrm{CO}_{2}\) are used when \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\) are produced?

We are given that \(0.400\, \mathrm{mol}\) of \(\mathrm{KO}_{2}\) reacts. From the balanced chemical equation, we have: $$\frac{3\, \mathrm{mol}\, \mathrm{O}_{2}}{4\, \mathrm{mol}\, \mathrm{KO}_{2}}$$ So, to find the moles of oxygen produced, we multiply the moles of \(\mathrm{KO}_{2}\) by the stoichiometric ratio: $$0.400\, \mathrm{mol}\, \mathrm{KO}_{2} \times \frac{3\, \mathrm{mol}\, \mathrm{O}_{2}}{4\, \mathrm{mol}\, \mathrm{KO}_{2}}$$ Let's calculate the moles of oxygen produced: $$0.400\, \mathrm{mol}\, \mathrm{KO}_{2} \times \frac{3}{4} = 0.300\, \mathrm{mol}\, \mathrm{O}_{2}$$ So, there are \(0.300\, \mathrm{mol}\) of \(\mathrm{O}_{2}\) produced.

We have to find the grams of \(\mathrm{KO}_{2}\) needed to form \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\). First, we need to find the moles of \(\mathrm{O}_{2}\) in \(7.50 \mathrm{~g}\). To do this, we will use the molar mass of oxygen: $$\frac{1\,\mathrm{mol}\, \mathrm{O}_{2}}{32.00\, \mathrm{g}\, \mathrm{O}_{2}}$$ Calculate the moles of oxygen: $$\frac{7.50\, \mathrm{g}\, \mathrm{O}_{2}}{32.00\, \mathrm{g/mol}} = 0.234\, \mathrm{mol}\, \mathrm{O}_{2}$$ Now, we will use the stoichiometric ratio from the balanced chemical equation: $$\frac{4\, \mathrm{mol}\, \mathrm{KO}_{2}}{3\, \mathrm{mol}\, \mathrm{O}_{2}}$$ Calculate the moles of \(\mathrm{KO}_{2}\) needed: $$0.234 \, \mathrm{mol}\, \mathrm{O}_{2} \times \frac{4}{3} = 0.312\, \mathrm{mol}\, \mathrm{KO}_{2}$$ Finally, we use the molar mass of \(\mathrm{KO}_{2}\) to find the grams of \(\mathrm{KO}_{2}\): $$\frac{1\, \mathrm{mol}\, \mathrm{KO}_{2}}{71.10\, \mathrm{g}\, \mathrm{KO}_{2}}$$ Calculate the grams of \(\mathrm{KO}_{2}\) needed: $$0.312\, \mathrm{mol}\, \mathrm{KO}_{2} \times 71.10\, \mathrm{g/mol} = 22.2\, \mathrm{g}\, \mathrm{KO}_{2}$$ So, \(22.2 \mathrm{~g}\) of \(\mathrm{KO}_{2}\) are needed to form \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\).

We know from (b) that \(0.234\, \mathrm{mol}\) of \(\mathrm{O}_{2}\) are produced from \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\). We will use the stoichiometric ratio from the balanced chemical equation to find the moles of \(\mathrm{CO}_{2}\): $$\frac{2\, \mathrm{mol}\, \mathrm{CO}_{2}}{3\, \mathrm{mol}\, \mathrm{O}_{2}}$$ Calculate the moles of \(\mathrm{CO}_{2}\) used: $$0.234\, \mathrm{mol}\, \mathrm{O}_{2} \times \frac{2}{3} = 0.156\, \mathrm{mol}\, \mathrm{CO}_{2}$$ Finally, we use the molar mass of \(\mathrm{CO}_{2}\) to find the grams of \(\mathrm{CO}_{2}\): $$\frac{1\, \mathrm{mol}\, \mathrm{CO}_{2}}{44.01\, \mathrm{g}\, \mathrm{CO}_{2}}$$ Calculate the grams of \(\mathrm{CO}_{2}\) used: $$0.156\, \mathrm{mol}\, \mathrm{CO}_{2} \times 44.01\, \mathrm{g/mol} = 6.86\, \mathrm{g}\, \mathrm{CO}_{2}$$ So, \(6.86 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) are used when \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\) are produced.