An electrolytic cell for aluminium carries a current of 150 kA for 6.0 hours at a voltage of 4.30 V…

Step 1. Charge. $Q = It = 150\,000 \times 6.0 \times 3600 = 3.24 \times 10^9\ \text{C}$.

Step 2. Aluminium. Moles of aluminium $= 0.92 \times 3.24 \times 10^9 / (3 \times 96485) = 1.030 \times 10^4\ \text{mol}$. The mass is $1.030 \times 10^4 \times 26.98 = 2.778 \times 10^5\ \text{g}$, which is $277.8\ \text{kg}$ or about $278\ \text{kg}$.

Step 3. Energy. $E = VQ = 4.30 \times 3.24 \times 10^9 = 1.393 \times 10^{10}\ \text{J}$. Divide by $3.6 \times 10^6$ to get $3870\ \text{kWh}$.

Step 4. For each kilogram. $3870 / 277.8 = 13.9\ \text{kWh/kg}$.

Step 5. Carbon. The carbon is $3/4$ of the moles of aluminium: $0.75 \times 1.030 \times 10^4 = 7723\ \text{mol}$, which is $7723 \times 12.01 = 92.8\ \text{kg}$.