Question

for the reaction 4 al(s) + 3 o₂(g) → 2 al₂o₃(s), when 1.5 moles of al are reacted with 3 moles of o₂, 0.70 mol of al₂o₃ are produced. what is the percent yield for this reaction?
○ 40%
○ 56%
○ 93%
○ 74%

Explanation:

Step1: Find limiting reactant

From reaction: \( 4 \text{Al}(s) + 3 \text{O}_2(g)
ightarrow 2 \text{Al}_2\text{O}_3(s) \)
Mole ratio of \( \text{Al} : \text{O}_2 \) is \( 4:3 \).
For 1.5 mol Al, required \( \text{O}_2 \): \( \frac{3}{4} \times 1.5 = 1.125 \) mol (available 3 mol, so Al is limiting).

Step2: Calculate theoretical yield

From Al (limiting), mole ratio \( \text{Al} : \text{Al}_2\text{O}_3 = 4:2 = 2:1 \).
Theoretical moles of \( \text{Al}_2\text{O}_3 \): \( \frac{1}{2} \times 1.5 = 0.75 \) mol.

Step3: Calculate percent yield

Percent yield \( = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100 = \frac{0.70}{0.75} \times 100 \approx 93.33\% \approx 93\% \).

Answer:

93% (corresponding to the option with 93%)