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 \(2 \mathrm{~mol}\) 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 \(\mathrm{CO}_{2}(g) .\) (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?

Given the products of the reaction are \(\mathrm{K}_{2} \mathrm{CO}_{3}(s)\) and \(\mathrm{O}_{2}(g)\), the unbalanced equation for the reaction between \(\mathrm{KO}_{2}(s)\) and \(\mathrm{CO}_{2}(g)\) is: \[\mathrm{KO}_{2}(s) + \mathrm{CO}_{2}(g) \rightarrow \mathrm{K}_{2} \mathrm{CO}_{3}(s) + \mathrm{O}_{2}(g).\] To balance the equation, we must take into account that \(2 \mathrm{~mol}\) of \(\mathrm{KO}_{2}(s)\) react with each mole of \(\mathrm{CO}_{2}(g)\), which means that we need 2 moles of \(\mathrm{KO}_{2}(s)\) for each mole of \(\mathrm{CO}_{2}(g)\). Therefore, the balanced equation is: \[2\mathrm{KO}_{2}(s) + \mathrm{CO}_{2}(g) \rightarrow \mathrm{K}_{2} \mathrm{CO}_{3}(s) + \frac{3}{2} \mathrm{O}_{2}(g).\]

The oxidation number for each atom involved in the reaction is as follows: - In \(\mathrm{KO}_{2}(s)\): K: +1, O: -1 - In \(\mathrm{CO}_{2}(g)\): C: +4, O: -2 - In \(\mathrm{K}_{2} \mathrm{CO}_{3}(s)\): K: +1, C: +4, O: -2 - In \(\mathrm{O}_{2}(g)\): O: 0 Notice that the oxidation number of oxygen changes from -1 in \(\mathrm{KO}_{2}\) to -2 in \(\mathrm{K}_{2}\mathrm{CO}_{3}\) and 0 in \(\mathrm{O}_{2}\). This means that oxygen is being reduced. On the other hand, the oxidation number of potassium remains constant, and so does that of carbon.

Given the mass of \(\mathrm{CO}_{2}(g)\) needed is \(18.0 \mathrm{~g}\), we will use stoichiometry to calculate the mass of \(\mathrm{KO}_{2}(s)\) required and the mass of \(\mathrm{O}_{2}(g)\) produced. First, we will find the moles of \(\mathrm{CO}_{2}(g)\): \[moles\; \mathrm{CO}_{2} = \frac{18.0 \mathrm{~g}}{44.01\frac{g}{mol}} = 0.4091\, mol\] Using the balanced equation, we can find the moles of \(\mathrm{KO}_{2}(s)\) needed: \[moles\; \mathrm{KO}_{2} = 0.4091\,mol\, \mathrm{CO}_{2} \times \frac{2\,mol\, \mathrm{KO}_{2}}{1\,mol\, \mathrm{CO}_{2}} = 0.8182\, mol\, \mathrm{KO}_{2}\] Now, we calculate the mass of \(\mathrm{KO}_{2}(s)\) required: \[mass\; \mathrm{KO}_{2} = 0.8182\,mol \times 71.1\frac{g}{mol} = 58.2\,g\] Next, we find the moles of \(\mathrm{O}_{2}(g)\) produced: \[moles\,\mathrm{O}_{2} = 0.4091\,mol\, \mathrm{CO}_{2} \times \frac{3}{2}\frac{mol\,\mathrm{O}_{2}}{1\,mol\, \mathrm{CO}_{2}} = 0.6136\,mol\, \mathrm{O}_{2}\] Finally, we calculate the mass of \(\mathrm{O}_{2}(g)\) produced: \[mass\; \mathrm{O}_{2} = 0.6136\,mol \times 32.00\frac{g}{mol} = 19.6\,g\] In conclusion, to consume \(18.0 \mathrm{~g} \mathrm{CO}_{2}(g)\), \(58.2\,\mathrm{g}\) of \(\mathrm{KO}_{2}(s)\) is needed, and \(19.6\,\mathrm{g}\) of \(\mathrm{O}_{2}(g)\) is produced during the reaction.