Through a solution containing 17 g of silver nitrate passed 3 liters of hydrogen sulfide

Through a solution containing 17 g of silver nitrate passed 3 liters of hydrogen sulfide, calculate the mass of the precipitate.

Given:
m (AgNO3) = 17 g
V (H2S) = 3 l

To find:
m (draft) -?

Decision:
1) 2AgNO3 + H2S => Ag2S ↓ + 2HNO3;
2) M (AgNO3) = Mr (AgNO3) = Ar (Ag) * N (Ag) + Ar (N) * N (N) + Ar (O) * N (O) = 108 * 1 + 14 * 1 + 16 * 3 = 170 g / mol;
M (Ag2S) = Mr (Ag2S) = Ar (Ag) * N (Ag) + Ar (S) * N (S) = 108 * 2 + 32 * 1 = 248 g / mol;
3) n (AgNO3) = m (AgNO3) / M (AgNO3) = 17/170 = 0.1 mol;
4) n (H2S) = V (H2S) / Vm = 3 / 22.4 = 0.13 mol;
5) n (Ag2S) = n (AgNO3) / 2 = 0.1 / 2 = 0.05 mol;
6) m (Ag2S) = n (Ag2S) * M (Ag2S) = 0.05 * 248 = 12.4 g.

Answer: The mass of Ag2S is 12.4 g.