A steady electric current is passed through molten \(\ce{MgCl2}\) for exactly 1.00 hour, producing 243 g of \(\ce{Mg}\) metal. If the same current is passed through molten \(\ce{AlCL3}\) for 1.00 hour, the mass of \(\ce{Al}\) metal produced is closest to

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The relevant equation for the reduction of magnesium ions:

$$ \ce{Mg^{2+} + 2e^- -> Mg} $$

Using stoichiometry, we can obtain the number of moles of electrons passed per hour.

$$ \pu{243 g }\ce{Mg} \cdot \frac{\pu{1 mol }\ce{Mg}}{\pu{24.3 g }\ce{Mg}} \cdot \frac{\pu{2 mol }\ce{e^-}}{\pu{1 mol }\ce{Mg}} $$ $$ = \pu{20 mol }\ce{ e^-} \text{ per hour} $$

The relevant equation for the reduction of aluminum ions:

$$ \ce{Al^{3+} + 3e^- -> Al} $$

Using a similar process as before:

$$ \pu{20 mol }\ce{e^-} \cdot \frac{\pu{1 mol }\ce{Al}}{\pu{3 mol }\ce{e^-}} \cdot \frac{\pu{27 g }\ce{Al}}{\pu{1 mol }\ce{Al}} $$ $$ = \boxed{\pu{180 g }\ce{Al}} $$