# 18 grams of butanal, containing 20% impurities, was acted upon by an excess of an ammoniacal

**18 grams of butanal, containing 20% impurities, was acted upon by an excess of an ammoniacal solution of silver oxide. Find the mass of sediment that forms.**

Let’s implement the solution:

1. Let us write down the equation according to the problem statement:

m = 18 g. 20% – impurities. X g -?

С4Н8О + Ag2O (NH4OH) = 2Ag + C3H7 – COOH – oxidation of butanal, obtained silver, butanoic (butyric) acid;

2. We make calculations:

M (C4H8O) = 68 g / mol;

M (Ag) = 107.8 g / mol;

m (C4H8O) = 18 * (1-0.20) = 14.4 g;

Y (C4H8O) = m / M = 14.4 / 68 = 0.2 mol.

3. Proportion:

0.2 mol (C4H8O) – X mol (Ag);

-1 mol -2 mol from here, X mol (Ag) = 0.2 * 2/1 = 0.4 mol.

4. Find the mass of the product:

m (Ag) = Y * M = 0.4 * 197.8 = 43.12 g.

Answer: silver was obtained with a mass of 43.12 g.