# Question #1724c

Jul 1, 2017

$n \left(R M g X\right) = 5.00 \cdot {10}^{-} 5 \text{ } m o l$

#### Explanation:

The amount of charge passed divided by the Faraday constant will give you the number of moles of electrons transferred:

$n = \frac{Q}{F}$

$F \approx 96500 \text{ } \frac{C}{m o l}$

$n \left({e}^{-}\right) = \frac{9.65}{96500} = {10}^{-} 4 \text{ } m o l$

Our starting material is comprised of Mg cations with a 2+ charge and Cl anions with a 1- charge.

$M g C {l}_{2} \equiv M {g}^{2 +} + 2 C {l}^{-}$

One mole of electrons will give 1/2 a mole of solid magnesium, because we need two electrons to reduce each magnesium cation:

$M {g}^{2 +} + 2 {e}^{-} \to M {g}_{\left(s\right)}$

So to get the amount of Mg created we halve the number of moles of electrons:

$n \left(M g\right) = \frac{1}{2} \cdot n \left({e}^{-}\right) = \frac{1}{2} \cdot {10}^{-} 4 = 5.00 \cdot {10}^{-} 5 \text{ } m o l$

The general formula for a grignard reagent is:

$R M g X$

This means that one mole of solid magnesium will give one mole of grignard reagent, so the number of moles of grignard is the same as the amount of solid magnesium we created:

$n \left(M g\right) = n \left(R M g X\right) = 5.00 \cdot {10}^{-} 5 \text{ } m o l$