11.2 L of propyne was passed through an ammonia solution containing 69.9 g of silver oxide.

11.2 L of propyne was passed through an ammonia solution containing 69.9 g of silver oxide. Determine the mass of precipitated sediment.

Given:
V (C3H4) = 11.2 L
m (Ag2O) = 69.9 g

To find:
m (draft) -?

Decision:
1) 2C3H4 + Ag2O => 2C3H3Ag ↓ + H2O;
2) n (C3H4) = V (C3H4) / Vm = 11.2 / 22.4 = 0.5 mol;
3) M (Ag2O) = Mr (Ag2O) = Ar (Ag) * N (Ag) + Ar (O) * N (O) = 108 * 2 + 16 * 1 = 232 g / mol;
4) n (Ag2O) = m (Ag2O) / M (Ag2O) = 69.9 / 232 = 0.3 mol;
5) n (C3H3Ag) = n (C3H4) = 0.5 mol;
6) M (C3H3Ag) = Mr (C3H3Ag) = Ar (C) * N (C) + Ar (H) * N (H) + Ar (Ag) * N (Ag) = 12 * 3 + 1 * 3 + 108 * 1 = 147 g / mol;
7) m (C3H3Ag) = n (C3H3Ag) * M (C3H3Ag) = 0.5 * 147 = 73.5 g.

Answer: The mass of C3H3Ag is 73.5 g.