Determine the theoretical yield, in L, of O2 produced from reacting 24.9 g KO2 (71.10 g/mol) with 3.90 L of CO2 at 273.15 K and 1.00 atm (a.k.a. STP). 4 KO2 (s) + 2 CO2 (g) → 2 K2CO3 (s) + 3 O2 (g) Molar volume of an ideal gas at STP = 22.4 L/mol 𝑃𝑉 = 𝑛𝑅𝑇

1. Calculate the moles of KO2:
\text{Moles of } KO2 = \frac{24.9}{71.10} \approx 0.350 \, \text{mol}

2. Calculate the moles of CO2:
n = \frac{1.00 \times 3.90}{0.0821 \times 273.15} \approx 0.174 \, \text{mol}

3. Determine the limiting reagent:
Since the ratio KO2:CO2 = 4:2 = 2:1, compare:
0.350 \, \text{mol KO2} \times \frac{2}{4} = 0.175 \, \text{mol CO2}
Given 0.174 mol CO2 < 0.175 mol, CO2 is the limiting reagent.

4. Calculate the moles of O2 produced:
Using CO2 as the limiting reagent:
\text{Moles of } O2 = 0.174 \, \text{mol CO2} \times \frac{3}{2} = 0.261 \, \text{mol}

5. Convert moles of O2 to volume at STP:
\text{Volume of } O2 = 0.261 \, \text{mol} \times 22.4 \, \frac{\text{L}}{\text{mol}} \approx 5.85 \, \text{L}

The theoretical yield of \( O_2 \) produced is 5.85 L.