# How many Mg^(2+) ions are present in 3.00 moles of MgCl_2?

Aug 6, 2018

$1.81 \text{ x "10^24 Mg^(2+) "ions}$

#### Explanation:

Step 1:
From the formula $M g C {l}_{2}$, we know that it's made from 1 $M {g}^{2 +}$ and 2 $C {l}^{-}$ ions

• In 1 mol of $M g C {l}_{2}$, there's 1 mol of $M {g}^{2 +}$

We can use this relationship to calculate the moles of $M {g}^{2 +} \text{ions}$ from the given 3.00 moles of $M g C {l}_{2}$.

Step 2:
Once we find the moles of $M {g}^{2 +}$, we can find the number of $M {g}^{2 +} \text{ions}$ using Avogadro's number.

• 1 mol $M {g}^{2 +}$ has $6.022$ x ${10}^{23}$ $M {g}^{2 +} \text{ions}$

Combining the 2 steps calculations:
$M {g}^{2 +} \text{ions"="3.00 mols " MgCl_2 xx ("1 mol " Mg^(2+))/("1 mol " MgCl_2)xx(6.022 " x " 10^23 Mg^(2+) ions)/("1 mol " Mg^(2+))=1.81" x "10^24 Mg^(2+) "ions}$