210 liters of ammonia reacted with sulfuric acid weighing 1.2 g.

210 liters of ammonia reacted with sulfuric acid weighing 1.2 g. Calculate the resulting solution of ammonium sulfate if the reaction proceeded with a yield of 78%

Given:
V (NH3) = 210 l
m (H2SO4) = 1.2 g
η ((NH4) 2SO4) = 78%

To find:
m ((NH4) 2SO4) -?

Decision:
1) 2NH3 + H2SO4 => (NH4) 2SO4;
2) M (H2SO4) = Mr (H2SO4) = Ar (H) * N (H) + Ar (S) * N (S) + Ar (O) * N (O) = 1 * 2 + 32 * 1 + 16 * 4 = 98 g / mol;
M ((NH4) 2SO4) = Mr ((NH4) 2SO4) = Ar (N) * N (N) + Ar (H) * N (H) + Ar (S) * N (S) + Ar (O) * N (O) = 14 * 2 + 1 * 8 + 32 * 1 + 16 * 4 = 132 g / mol;
3) n (NH3) = V (NH3) / Vm = 210 / 22.4 = 9.375 mol;
4) n (H2SO4) = m (H2SO4) / M (H2SO4) = 1.2 / 98 = 0.012 mol;
5) n ((NH4) 2SO4) = n (H2SO4) = 0.012 mol;
6) m theor. ((NH4) 2SO4) = n ((NH4) 2SO4) * M ((NH4) 2SO4) = 0.012 * 132 = 1.584 g;
7) m practical. ((NH4) 2SO4) = η ((NH4) 2SO4) * m theory. ((NH4) 2SO4) / 100% = 78% * 1.584 / 100% = 1.24 g.

Answer: The practical mass (NH4) 2SO4 is 1.24 g.