Find A, B and C in the following reactions:
$$NH_3 + A + CO_2 \to (NH_4)_2CO_3$$
$$(NH_4)_2CO_3 + H_2O + B \to NH_4HCO_3$$
$$NH_4HCO_3 + NaCl \to NH_4Cl + C$$

These reactions are part of the Solvay process for manufacturing sodium bicarbonate. Let us balance each reaction to identify A, B, and C.

Reaction 1: $$2NH_3 + H_2O + CO_2 \to (NH_4)_2CO_3$$. Here, two moles of ammonia react with water and carbon dioxide to form ammonium carbonate. So $$A = H_2O$$.

Reaction 2: $$(NH_4)_2CO_3 + H_2O + CO_2 \to 2NH_4HCO_3$$. Ammonium carbonate reacts with water and carbon dioxide to form ammonium bicarbonate. So $$B = CO_2$$.

Reaction 3: $$NH_4HCO_3 + NaCl \to NH_4Cl + NaHCO_3$$. This is a double displacement reaction where ammonium bicarbonate reacts with sodium chloride to form ammonium chloride and sodium bicarbonate. So $$C = NaHCO_3$$.

Therefore, $$A = H_2O$$, $$B = CO_2$$, and $$C = NaHCO_3$$.