Hydrogen sulfide with a volume of 4.48 liters was passed through a solution of sodium hydroxide with a volume of 144.1

Hydrogen sulfide with a volume of 4.48 liters was passed through a solution of sodium hydroxide with a volume of 144.1 ml with w (NaOH) = 10% and p = 1.11 g / ml. What salt and what mass was obtained?

Given:
V solution (NaOH) = 144.1 ml
ω (NaOH) = 10%
ρ solution (NaOH) = 1.11 g / ml
V (H2S) = 4.48 L

To find:
m (salt) -?

Decision:
1) m solution (NaOH) = ρ solution (NaOH) * V solution (NaOH) = 1.11 * 144.1 = 159.951 g;
2) m (NaOH) = ω (NaOH) * m solution (NaOH) / 100% = 10% * 159.951 / 100% = 15.995 g;
3) n (NaOH) = m (NaOH) / M (NaOH) = 15.995 / 40 = 0.4 mol;
4) n (H2S) = V (H2S) / Vm = 4.48 / 22.4 = 0.2 mol;
5) 2NaOH + H2S => Na2S + 2H2O;
6) n (Na2S) = n (H2S) = 0.2 mol;
7) m (Na2S) = n (Na2S) * M (Na2S) = 0.2 * 78 = 15.6 g.

Answer: The mass of Na2S is 15.6 g.