In 735 mg of a 20% sulfuric acid solution, 30 ml of ammonia was dissolved. Calculate the mass of the salt formed.

Given:
m solution (H2SO4) = 735 mg = 0.735 g
ω (H2SO4) = 20%
V (NH3) = 30 ml = 0.03 l

Find:
m (salt) -?

1) m (H2SO4) = ω * m solution / 100% = 20% * 0.735 / 100% = 0.147 g;
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;
3) n (H2SO4) = m / M = 0.147 / 98 = 0.0015 mol;
4) n (NH3) = V / Vm = 0.03 / 22.4 = 0.0013 mol;
5) H2SO4 + NH3 => NH4HSO4;
6) n (NH4HSO4) = n (NH3) = 0.0013 mol;
7) M (NH4HSO4) = Mr (NH4HSO4) = Ar (N) * N (N) + Ar (H) * N (H) + Ar (S) * N (S) + Ar (O) * N (O ) = 14 * 1 + 1 * 5 + 32 * 1 + 16 * 4 = 115 g / mol;
8) m (NH4HSO4) = n * M = 0.0013 * 115 = 0.15 g.

Answer: The mass of NH4HSO4 is 0.15 g.