Question

Aluminum oxide (Al2O3) is produced according to the following equation.

4 Al(s) + 3 O2(g) → 2 Al2O3(s)

If the reaction occurs with an 82.4% yield, what mass of aluminum should be reacted with excess oxygen to produce 45.0 grams of Al2O3?

a. 54.6 g Al

b. 37.9 g Al

c. 35.1 g Al

d. 23.8 g Al

e. 28.9 g Al

How do we solve this problem

Answer 1

4 Al(s) + 3 O2(g) → 2 Al2O3(s)

Molar mass of Al = 26.98 g/mol

Molar mass of O2 = 31.99 g/mol

Molar mass of Al2O3 = 101.96 g/mol

From above balanced reaction:

4 x 26.98 = 107.92 g of Al produces 2 x 101.96 = 203.92 g of Al2O3

Hence, amount 203.92 g of Al2O3 is 100% yield.

But given condition is only for 82.4% yield, so mass of Al2O3 associated with 82.4% yield

= 82.4 x 203.92 / 100

= 168.03 g of Al2O3

Now, as 168.03 g of Al2O3 requires 107.92 g of Al

So, Al required to produce 45.0 g of Al2O3 = 45.0 x 107.92 / 168.03

= 28.9 g of Al