Question #37918

Feb 23, 2018

$1.8496$ moles of electrons

Explanation:

Writing down the entire equation with electrons included will help solve the problem:

$F e C {l}_{3} \left(a q\right) + 3 {e}^{-} \rightarrow F e \left(s\right) + 3 C {l}^{-} \left(a q\right)$

There are three electrons on the left side because the iron in iron trichloride has an oxidation state of $+ 3$, and it gets reduced to an oxidation of zero in its elemental form. The three electrons balance the opposing charges on both sides. Notice how for every mole of $F e C {l}_{3}$ is reduced, three moles of electrons are required.

Besides the $F e C {l}_{3}$ to electron mole ratio, you need to know the molar mass of $F e C {l}_{3}$ which is 162.195g g/mol. (You can do this this using the periodic table.)

Now, using a T-chart, you can solve the problem.

$100 g F e C {l}_{3} \cdot \left(\frac{1 m o l F e C {l}_{3}}{162.195 g}\right) \cdot \left(\frac{3 m o l {e}^{-}}{1 m o l F e C {l}_{3}}\right)$

Multiplying it all out, you will receive an answer of $1.8496$. Checking the units, you will notice that grams $F e C {l}_{3}$ cancel out in the first conversion, and moles $F e C {l}_{3}$ cancel out in the second conversion, leaving just moles electrons.