What is the percent yield of the reaction below if 84.0 grams of Al2O3(s) is recovered from a reaction whose theoretical yield of Al2O3(s) is 104 grams? 4 Al(s) + 3 O2(g) → 2 Al2O3(s)

80.8%

Explanation:

Let's consider the following balanced equation.

4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

The mass obtained of Al₂O₃ (experimental yield) is 84.0 g. The theoretical yield of Al₂O₃ is 104 g. We can calculate the percent yield of Al₂O₃ using the following expression.

%yield = (experimental yield / theoretical yield) × 100%

%yield = (84.0 g / 104 g) × 100% = 80.8%

Percent Yield = 80.8%

Explanation:

We can find the percent yield of a reaction using the equation:

Percent yield = Actual yield (g) / Theoretical Yield (g) * 100

Where Actual yield is the amount of product produced (84.0g)

And theoretical yield is the mass produced assuming a 100% of product (104.0g)

Replacing the computed values:

Percent yield = 84.0g / 104.0g * 100

Percent Yield = 80.8%

dynamic equilibrium is the answer.