# How do you balance (NH_4)_2CO_3 + CuCl_2 -> NH_4Cl + CuCO_3?

We see one equivalent of ${\text{Cu}}^{2 +}$ (copper(II)) on each side, and one equivalent of ${\text{CO}}_{3}^{2 -}$ (carbonate) on each side.
However, we only have one equivalent of ${\text{NH}}_{4}^{+}$ (ammonium) on the right side and one equivalent of ${\text{Cl}}^{-}$ (chloride) on the right side, but two equivalents of each on the left side. (The subscripts outside parentheses correspond to the whole polyatomic ion.)
Therefore, all we need is a $2$ on the $\textcolor{w h i t e}{\text{NH"_4"Cl}}$. As a result, you get (highlight the answer to check):
color(white)(("NH"_4)_2"CO"_3 + "CuCl"_2 -> 2"NH"_4"Cl" + "CuCO"_3)