Question

47 g of impure Al₂O₃ is reduced electrolytically to give aluminium and oxygen. Calculate the amount of aluminium produced.

Answer 1

Electrolytic reduction of Aluminium takes place as follows

$$\begin{gathered} 2{\mathrm{Al}}_2{\mathrm{O}}_3\stackrel{\mathrm{electricity}}{\to }4\mathrm{Al}+3{\mathrm{O}}_2 \\ (\mathrm{Aluminium}\mathrm{oxide})\left(\mathrm{Aluminium}\right)\left(\mathrm{Oxygen}\right) \end{gathered}$$

• Gram atomic weight of Aluminium= 27 g mol^-1
• Gram atomic weight of Oxygen= 16 g mol^-1

Step-1: Calculation of the molecular weight

The gram molecular weight of Aluminium oxide = 2(Gram atomic weight of Aluminium) + 3(Gram atomic weight of Oxygen)

Gram molecular weight of Aluminium oxide = 2(27) +3(16)

The gram molecular weight of Aluminium oxide= 102 g mol^-1

The gram atomic weight of Aluminium= 27 g mol^-1

Step-1: Calculation of the amount of aluminium produced

In a balanced chemical equation, 2 moles of aluminium oxide undergo reduction to form 4 moles of aluminium.

• As 1 mole of a substance= gram molecular weight of a substance (if the substance is a molecule). Therefore, 2 mole of aluminium oxide= 2×102 = 204 g
• As 1 mole of a substance= gram atomic weight of a substance (if the substance is an element). Therefore, 4 mole of aluminium = 4 ×27= 108 g

Therefore,

204 g of impure aluminium oxide produces aluminium= 108 g

1g of impure aluminium oxide produces aluminium= $\frac{108}{204}$ g

47g of impure aluminium oxide produces aluminium= $\frac{108}{204}\times 47=24.88\approx 25$ g

Therefore, the amount of aluminium produced from 47 g of impure Al₂O₃ is 25 g.