Through 300 g of 20% sodium hydroxide solution, 44.8 liters of hydrogen sulfide are passed. Mass of salt.

Given:
m solution (NaOH) = 300 g
ω (NaOH) = 20%
V (H2S) = 44.8 L

To find:
m (salt) -?

Decision:
1) M (NaOH) = Mr (NaOH) = Ar (Na) * N (Na) + Ar (O) * N (O) + Ar (H) * N (H) = 23 * 1 + 16 * 1 + 1 * 1 = 40 g / mol;
2) m (NaOH) = ω (NaOH) * m solution (NaOH) / 100% = 20% * 300/100% = 60 g;
3) n (NaOH) = m (NaOH) / M (NaOH) = 60/40 = 1.5 mol;
4) n (H2S) = V (H2S) / Vm = 44.8 / 22.4 = 2 mol;
5) H2S + NaOH => NaHS + H2O;
6) M (NaHS) = Mr (NaHS) = Ar (Na) * N (Na) + Ar (H) * N (H) + Ar (S) * N (S) = 23 * 1 + 1 * 1 + 32 * 1 = 56 g / mol;
7) n (NaHS) = n (NaOH) = 1.5 mol;
8) m (NaHS) = n (NaHS) * M (NaHS) = 1.5 * 56 = 84 g.

Answer: The mass of NaHS is 84 g.