Question

If 16.0 grams of aluminum oxide were actually produced, what is the percent yield of the reaction below given that you start with 10.0 g of Al and 19.0 grams of O 2?

Answer 1

Answer: The percent of yield is 50.03%.

Explanation:

First, we need to balance the equation:

$4Al + 3O_{2}$ ⇒ $2Al_{2}O_{3}$

We need to remember that the chemical equations are written in moles, so we have to convert the amounts in grams to moles, using the molecular weight of every compound: Al2O3 (101.96 g/mol), Al (29.98 g/mol) and O2 (31.99 g/mol).

In consequence, the amounts in moles of every compound will be:

$16 gAl_{2}O_{3} * \frac{1 mol}{101.96 g} =0.157 mol Al_{2}O_{3}$

$10 g Al * \frac{1mol}{29.98 g}= 0.333 mol Al$

$19 g O_{2}*\frac{1 mol}{31.99 g} =0.594 mol O_{2}$

Now, we have to find out which is the limit reagent or in other words, which of the reagents will be consumed first, taking into account the stoichiometric ratio of the balanced equation:

$0.333 mol Al * \frac{2 mol Al_{2}O_{3}}{4 mol Al} =0.1665 mol Al_{2}O_{3}$

$0.594 mol O_{2} * \frac{2 mol Al_{2}O_{3}}{3 mol O_{2}} =0.396 mol Al_{2}O_{3}$

As you can see, the maximum amount (theoretically) of Al2O3 that can be produced is 0.1665 mol.

Finally, we have to use the yield formula to calculate the percent yield of the reaction:

$Percent of yield = \frac{actual yield}{theoretical yield} * 100 = \frac{0.157 mol Al_{2}O_{3}}{0.1665 mol Al_{2}O_{3}} * 100 = 94.25$

Therefore, the percent of yield is 50.03%.