# 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.

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