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

1 Answer
Mar 12, 2016

Really, if you examine each molecule and atom as a unit, then almost everything is balanced already.

We see one equivalent of #"Cu"^(2+)# (copper(II)) on each side, and one equivalent of #"CO"_3^(2-)# (carbonate) on each side.

However, we only have one equivalent of #"NH"_4^(+)# (ammonium) on the right side and one equivalent of #"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 #color(white)("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)#