Question

This question involves two calculations. The answer to the first part will be
used for calculating the second part.
Aluminum powder, Al, will react and burn with oxygen gas, O2, to produce
aluminum oxide, Al2O3, according to the following balanced equation:
4 Al + 302
+ 2Al2O3
a) If 85.1 grams of aluminum react with excess oxygen in the air, how many
grams of aluminum oxide, Al2O3, will theoretically be produced?
b) If after the reaction is completed, 116.9 g of Al2O3 were actually
recovered and measured, what is the percent yield of the reaction?

Answer 1

Answer :

(a) The mass of $Al_2O_3$ produced is, 15.2 grams.

(b) The percent yield of the reaction is, 72.5 %

Explanation :

Part (a) :

Given,

Mass of $Al$ = 85.1 g

Molar mass of $Al$ = 27 g/mol

First we have to calculate the moles of $Al$

$\text{Moles of }Al=\frac{\text{Given mass }Al}{\text{Molar mass }Al}=\frac{85.1g}{27g/mol}=3.15mol$

Now we have to calculate the moles of $Al_2O_3$

The balanced chemical equation is:

$4Al+3O_2\rightarrow 2Al_2O_3$

From the reaction, we conclude that

As, 4 moles of $Al$ react to give 2 moles of $Al_2O_3$

So, 3.15 moles of $Al$ react to give $\frac{2}{4}\times 3.15=1.58$ mole of $Al_2O_3$

Now we have to calculate the mass of $Al_2O_3$

$\text{ Mass of }Al_2O_3=\text{ Moles of }Al_2O_3\times \text{ Molar mass of }Al_2O_3$

Molar mass of $Al_2O_3$ = 102 g/mole

$\text{ Mass of }Al_2O_3=(1.58moles)\times (102g/mole)=161.2g$

Therefore, the mass of $Al_2O_3$ produced is, 161.2 grams.

Part (b) :

Now we have to calculate the percent yield of the reaction.

$\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield = 116.9 g

Theoretical yield = 161.2 g

Now put all the given values in this formula, we get:

$\text{Percent yield}=\frac{116.9g}{161.2g}\times 100=72.5\%$

Therefore, the percent yield of the reaction is, 72.5 %