Question

1. what is the percent yield of the reaction below if 84.0 grams of $\text{al}_2\text{o}_3(s)$ is recovered from a reaction whose theoretical yield of $\text{al}_2\text{o}_3(s)$ is 104 grams?

$4\text{al}(s) + 3\text{o}_2(g) \
ightarrow 2\text{al}_2\text{o}_3(s)$

Explanation:

Step1: Recall percent yield formula

Percent yield = $\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

Step2: Substitute given values

Actual yield = 84.0 g, Theoretical yield = 104 g
Percent yield = $\frac{84.0}{104} \times 100\%$

Step3: Calculate the result

$\frac{84.0}{104} \approx 0.8077$, so $0.8077 \times 100\% = 80.8\%$ (rounded to three significant figures)

Answer:

80.8%