Potassium superoxide, \(\mathrm{KO}_{2},\) is often used in oxygen masks (such as those used by firefighters) because \(\mathrm{KO}_{2}\) reacts with \(\mathrm{CO}_{2}\) to release molecular oxygen. Experiments indicate that 2 mol of \(\mathrm{KO}_{2}(s)\) react with each mole of \(\mathrm{CO}_{2}(g) .\) (a) The products of the reaction are \(\mathrm{K}_{2} \mathrm{CO}_{3}(s)\) and \(\mathrm{O}_{2}(g) .\) Write a balanced equation for the reaction between \(\mathrm{KO}_{2}(s)\) and \(\mathrm{CO}_{2}(g) .(\mathbf{b})\) Indicate the oxidation number for each atom involved in the reaction in part (a). What elements are being oxidized and reduced? (c) What mass of \(\mathrm{KO}_{2}(s)\) is needed to consume \(18.0 \mathrm{~g} \mathrm{CO}_{2}(g)\) ? What mass of \(\mathrm{O}_{2}(g)\) is produced during this reaction?

We are given that 2 moles of \(\mathrm{KO}_{2}\) react with 1 mole of \(\mathrm{CO}_{2}\), and the products are \(\mathrm{K}_{2}\mathrm{CO}_{3}\) and \(\mathrm{O}_{2}\). Based on these initial reactants and products, the equation would look like this: \[2 \mathrm{KO}_{2}(s) + \mathrm{CO}_{2}(g) \rightarrow \mathrm{K}_{2}\mathrm{CO}_{3}(s) + \mathrm{O}_{2}(g)\] This equation is already balanced, as there are the same number of each type of atom on both sides of the equation.

In order to identify the elements that are being oxidized and reduced, we first need to determine the oxidation numbers of each element. For \(\mathrm{KO}_{2}\): K: +1 (potassium is an alkali metal and always has an oxidation number of +1) O: -1/2 (oxygen in superoxides has an oxidation number of -1/2) For \(\mathrm{CO}_{2}\): C: +4 (carbon in \(\mathrm{CO}_{2}\) has an oxidation number of +4) O: -2 (oxygen in most compounds has an oxidation number of -2) For \(\mathrm{K}_{2}\mathrm{CO}_{3}\): K: +1 C: +4 O: -2 For \(\mathrm{O}_{2}\): O: 0 (in its elemental form, oxygen has an oxidation number of 0)

Using the oxidation numbers we found in Step 2, we can now identify the elements being oxidized and reduced. Potassium's oxidation number remains the same, so it is not being oxidized or reduced. Oxygen's oxidation number changes from -1/2 in \(\mathrm{KO}_{2}\) to 0 in \(\mathrm{O}_{2}\), so it is being reduced. Carbon's oxidation number remains the same, so it is not being oxidized or reduced.

We are given that 18.0 g of \(\mathrm{CO}_{2}\) needs to react with \(\mathrm{KO}_{2}\). We can use stoichiometry to find the mass of \(\mathrm{KO}_{2}\) needed. First, convert the mass of \(\mathrm{CO}_{2}\) to moles: \[18.0 \, \text{g} \, \mathrm{CO}_{2} \cdot \frac{1 \, \text{mol} \, \mathrm{CO}_{2}}{44.01 \, \text{g} \, \mathrm{CO}_{2}} = 0.409 \, \text{mol}\, \mathrm{CO}_{2} \] Now, using the balanced equation from Step 1, we can find the moles of \(\mathrm{KO}_{2}\) needed: \[0.409 \, \text{mol}\, \mathrm{CO}_{2} \cdot \frac{2 \, \text{mol} \, \mathrm{KO}_{2}}{1 \, \text{mol} \, \mathrm{CO}_{2}} = 0.818 \, \text{mol}\, \mathrm{KO}_{2} \] Finally, convert the moles of \(\mathrm{KO}_{2}\) to grams: \[0.818 \, \text{mol}\, \mathrm{KO}_{2} \cdot \frac{71.10 \, \text{g} \, \mathrm{KO}_{2}}{1 \, \text{mol} \, \mathrm{KO}_{2}} = 58.18\text{g}\, \mathrm{KO}_{2} \] Therefore, 58.18 g of \(\mathrm{KO}_{2}\) is needed to react with 18.0 g of \(\mathrm{CO}_{2}\).

Using stoichiometry and the balanced equation, we can now find the mass of \(\mathrm{O}_{2}\) produced. Convert the moles of \(\mathrm{CO}_{2}\) that reacted to moles of produced \(\mathrm{O}_{2}\): \[0.409 \, \text{mol}\, \mathrm{CO}_{2} \cdot \frac{1 \, \text{mol} \, \mathrm{O}_{2}}{1 \, \text{mol} \, \mathrm{CO}_{2}} = 0.409 \, \text{mol}\, \mathrm{O}_{2} \] Finally, convert the moles of \(\mathrm{O}_{2}\) to grams: \[0.409 \, \text{mol}\, \mathrm{O}_{2} \cdot \frac{32.00 \, \text{g} \, \mathrm{O}_{2}}{1 \, \text{mol} \, \mathrm{O}_{2}} = 13.09\text{g}\, \mathrm{O}_{2} \] Therefore, 13.09 g of \(\mathrm{O}_{2}\) is produced during the reaction.