# How do you balance AgNO_3 +MgCl_2 → AgCl + Mg(NO_3)_2?

$2 {\text{AgNO"_3 +"MgC"l_2 → 2"AgC"l + "Mg(NO"_3")}}_{2}$
On the RHS, we have two nitrate ions. Therefore, we need $2$ mols of ${\text{AgNO}}_{3}$ to balance them. We now have two mols of ${\text{Ag}}^{+}$ on the LHS but only one mol on the RHS. Thus we also need $2$ mols $\text{AgC} l$.