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(D) The reduction reaction for the production of $Al$ from molten $Al_2O_3$ is given by: $Al^{3+} + 3e^- \rightarrow Al$ From the stoichiometry of the

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$0.3$

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(D) The reduction reaction for the production of $Al$ from molten $Al_2O_3$ is given by: $Al^{3+} + 3e^- ightarrow Al$ From the stoichiometry of the reaction,$1 \ mol$ of $Al$ $(27 \ g)$ requires $3 \ F$ of electricity. Given mass of $Al = 2.7 \ g$. Number of moles of $Al = \frac{2.7 \ g}{27 \ g/mol} = 0.1 \ mol$. Since $1 \ mol$ of $Al$ requires $3 \ F$,then $0.1 \ mol$ of $Al$ requires $0.1 	imes 3 = 0.3 \ F$. Therefore,the required electricity is $0.3 \ F$.

How much electricity in terms of Faraday is required to produce $2.7 \ g$ of $Al$ from molten $Al_2O_3$? (Atomic mass of $Al = 27 \ u$)

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