Question

What is the theoretical yield of aluminum that can be produced by the reaction of 60.0 g of aluminum oxide with 30.0 g of carbon according to the following chemical equation? ALLO, + 3C SH 2Al + 3CO
a. 31.8 g
b. 30g
c. 101.2 g
d. 45 g
e. 7.9 g

Answer 1

Answer: 31.8 g

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} Al_2O_3=\frac{60.0g}{102g/mol}=0.59moles$

$\text{Moles of} C=\frac{30.0g}{12g/mol}=2.5moles$

$Al_2O_3+3C\rightarrow 2Al+3CO$  

According to stoichiometry :

1 mole of $Al_2O_3$ require 3 moles of $C$

Thus 0.59 moles of $Al_2O_3$ will require=$\frac{3}{1}\times 0.59=1.77moles$  of $C$

Thus $Al_2O_3$ is the limiting reagent as it limits the formation of product and $C$ is the excess reagent as it is present in more amount than required.

As 1 mole of $Al_2O_3$ give = 2 moles of $Al$

Thus 0.59 moles of $Al_2O_3$ give =$\frac{2}{1}\times 0.59=1.18moles$  of $Al$

Mass of $Al=moles\times {\text {Molar mass}}=1.18moles\times 27g/mol=31.8g$

Thus 31.8 g of $Al$ will be produced from the given masses of both reactants.