Question

# 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

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## Answer to a math question 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

Hank
4.8
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.

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