If 2.52g Al reacts with an excess of Fe2O3 and recover 3.89 of alumina, Al2O3, what is the percent yield?

Answer 1

Hi, Caitlyn!

Percent yield is always (amount produced / amount expected) * 100%

To find the amount expected, we'll have to first write the equation for this reaction. We know that the reactants are Al and Fe₂O₃ and that one of the products is Al₂O₃. Fe is not represented in the products yet, though, so it must be a second product (each element has to appear on both sides of the reaction). As a starting point, we get:

Al + Fe₂O₃ --> Fe + Al₂O₃

We're not done yet, though, as we need to balance the equation. There are 2 Fe in the reactants but only 1 in the products. I can balance this by adding a coefficient of 2 to the Fe in the products. Similarly, there are 2 Al in the products but only 1 in the reactants. I can balance this by adding a coefficient of 2 to the Al in the reactants. Therefore, the balanced equation is:

2Al + Fe₂O₃ --> 2Fe + Al₂O₃

The problem gives us grams of Al and asks us to find the expected grams of Al₂O₃ that would result. To do this, I'll need to use dimensional analysis to move through units until I get to the desired end of grams of Al₂O₃. Starting at grams of Al, I can do this by moving to moles of Al, then to moles of Al₂O₃ by using the coefficients in the chemical equation, then to grams of Al₂O₃. As I do this, I'll always put the unit I'm moving towards as the numerator in my comparison point and the unit I'm moving away from in the denominator.

How do I know how many grams of Al there are in a mole of Al? By using the atomic mass from the periodic table. Similarly, I'll find the grams of Al₂O₃ in a mole of Al₂O₃ by adding twice the atomic mass of Al to three times the atomic mass of O.

2.52 g Al (1 mole Al / 26.98 g Al) (1 mole Al₂O₃ / 2 mole Al) (101.96 g Al₂O₃ / 1 mole Al₂O₃) = 4.76 g Al₂O₃

Now that we know the expected amount of Al₂O₃ that would be generated, we can plug it into the equation for percent yield along with the amount produced, which was provided in the question.

Percent yield = (amount produced / amount expected) * 100%

Percent yield = (3.89 g Al₂O₃ / 4.76 g Al₂O₃) * 100% = 81.72%

I hope this helped!