Question

In Part A, we saw that the theoretical yield of aluminum oxide is 0.800 mol . Calculate the percent yield if the actual yield of aluminum oxide is 0.456 mol .

Answer 1

Percent Yield is the actual practical yield of a certain chemical reaction. It is always less than the theoretical yield calculated due to human handling and error.

Percent Yield = (Actual yield / theoretical yiield) * 100 = (0.456 / 0.800) * 100 = 57%

The yield here is small which means a bad handling and high error.

Answer 2

Answer:

The percent yield is 58.12 %.

Explanation:

Theoretical yield of aluminum oxide = 0.800 mol

Theoretical amount of aluminum oxide = 0.800 mol × 102 g/mol = 81.6 g

Actual or experimental yield of aluminum oxide = 0.465 mol

Experimental amount of aluminum oxide  = 0.465 mol × 102 g/mol = 47.43 g

$\% yield=frac{\text{Experimental yield}}{\text{Actual yield}}\times 100$

$\%=\frac{47.43 g}{81.6 g}\times 100=58.12\%$

The percent yield is 58.12 %.