Ammonia was mixed with a volume of 5.6 l with hydrogen chloride in a volume of 6.72 l.

Ammonia was mixed with a volume of 5.6 l with hydrogen chloride in a volume of 6.72 l. Determine the volume of gas remaining and the mass of salt formed.

Given:
V (NH3) = 5.6 L
V (HCl) = 6.72 L

To find:
V (remainder) -?
m (salt) -?

Decision:
1) NH3 + HCl => NH4Cl;
2) V react. (HCl) = V (NH3) = 5.6 L;
3) V rest. (HCl) = V (HCl) – V re. (HCl) = 6.72 – 5.6 = 1.12 L;
4) n (NH3) = V (NH3) / Vm = 5.6 / 22.4 = 0.25 mol;
5) n (NH4Cl) = n (NH3) = 0.25 mol;
6) M (NH4Cl) = Mr (NH4Cl) = Ar (N) * N (N) + Ar (H) * N (H) + Ar (Cl) * N (Cl) = 14 * 1 + 1 * 4 + 35.5 * 1 = 53.5 g / mol;
7) m (NH4Cl) = n (NH4Cl) * M (NH4Cl) = 0.25 * 53.5 = 13.38 g.

Answer: The volume of HCl is 1.12 liters; the mass of NH4Cl is 13.38 g.