Electrochemistry Question 3

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

27.0 g

120. g

180. g

270. g

Approach

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}}