Question

42.4g of Al react with excess O₂ and 67.3g of Al₂O3 are produced, what was the percent yield of Al₂83?

Note: the balanced equation is: 4 A1+3 0₂ 2 Al₂O3

Answer 1

84.5%

Step-by-step explanation:

To calculate the percent yield, we need to first calculate the theoretical yield, which is the amount of product that would be produced if the reaction went to completion.

The balanced chemical equation is:

4 Al + 3 O₂ → 2 Al₂O₃

From the equation, we can see that 4 moles of Al react with 3 moles of O₂ to produce 2 moles of Al₂O₃. Therefore, we need to find the moles of Al in the reaction:

42.4 g Al × (1 mol Al / 26.98 g Al) = 1.57 mol Al

Using the mole ratio from the balanced equation, we can calculate the theoretical yield of Al₂O₃:

1.57 mol Al × (2 mol Al₂O₃ / 4 mol Al) × (101.96 g Al₂O₃ / 1 mol Al₂O₃) = 79.6 g Al₂O₃

Now we can calculate the percent yield:

(actual yield / theoretical yield) × 100%

The actual yield is given as 67.3 g Al₂O₃, so:

(67.3 g Al₂O₃ / 79.6 g Al₂O₃) × 100% = 84.5%

Therefore, the percent yield of Al₂O₃ is 84.5%.

ALLEN