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$\mathrm{Al}_2 \mathrm{O}_3$ is reduced by electrolysis at low potentials and high currents. If $4.0 \times 10^4$ amperes of current is passed through molten $\mathrm{Al}_2 \mathrm{O}_3$ for 6 hours, what mass of aluminium is produced ? (Assume 100\% current efficiency, at. mass of $\mathrm{Al}=27 \mathrm{~g} \mathrm{~mol}^{-1}$ ) -
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Verified Answer
The correct answer is:
$8.1 \times 10^4 \mathrm{~g}$
$$
\begin{aligned}
\mathrm{W} & =\frac{\mathrm{E}}{96500} \times \mathrm{I} \times \mathrm{t} \\
\mathrm{W} & =\frac{9}{96500} \times 4.0 \times 10^4 \times 6 \times 3600 \\
& =8.1 \times 10^4 \mathrm{~g}
\end{aligned}
$$
\begin{aligned}
\mathrm{W} & =\frac{\mathrm{E}}{96500} \times \mathrm{I} \times \mathrm{t} \\
\mathrm{W} & =\frac{9}{96500} \times 4.0 \times 10^4 \times 6 \times 3600 \\
& =8.1 \times 10^4 \mathrm{~g}
\end{aligned}
$$
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