A factory produces 40 kg of calcium in two hours by electrolysis. How much aluminum can be produced by same current in 2 hours if current efficiency is 50%?

$\begin{array}{l}2\mathrm{F}\left(\text{ or }\right)2\times 96500\mathrm{C}\to 40\mathrm{g}\\ \left(\mathrm{x}\right)?\to 40\times {10}^3\mathrm{g}\\ \mathrm{x}=\frac{2\times 96500\times 40\times {10}^3}{40}=2\times 965\times {10}^5\mathrm{C}\end{array}$
If current efficiency is 50% then $2\times 965\times {10}^5\times \frac{50}{100}\Rightarrow 965\times {10}^5\mathrm{C}$
For aluminium 
$\begin{array}{l}3\mathrm{F}\left(\text{ or }\right)3\times 96500\mathrm{C}\to 27\mathrm{g}\\ 965\times {10}^5\mathrm{C}\to ?\left(\mathrm{y}\right)\\ \mathrm{y}=\frac{27\times 965\times {10}^5}{3\times 965\times {10}^2}=9\times {10}^3\mathrm{g}(\text{ or) }9\mathrm{kg}\end{array}$