# What is the weight in grams of silver displaced by 0.1 mole of magnesium?

Mar 24, 2018

$215.736$ grams.

#### Explanation:

Magnesium metal displaces silver from an aqueous solution according to the formula
$M g \left(s\right) + 2 A {g}^{+} \left(a q\right) \to M {g}^{2 +} \left(a q\right) + 2 A g \left(s\right)$

Thus $\frac{n \left(A g\right)}{n \left(M g\right)} = \frac{2}{1}$
$n \left(A g\right) = n \left(M g\right) \cdot \frac{n \left(M g\right)}{n \left(A g\right)} = 0.1 \cdot \frac{2}{1} = 0.2 \cdot \text{mol}$

$M \left(A g\right) = 107.868 g \cdot m o {l}^{- 1}$ from the periodic table. Therefore
$m \left(A g\right) = n \left(A g\right) \cdot M \left(A g\right) = 215.736 \cdot g$

Note how the charge conserves on either side of the equation. The conservation of charge can be something helpful to refer to in case you forget the ratio during an exam. As an alkaline-earth metal (i.e., of the second column of the periodic table from the left) magnesium ion would have a charge of $+ 2$ in solution. Still, you need to memorize that most silver ions in an aqueous solution have oxidation state $+ 1$.