What is the mass of the released silver, which is formed during the oxidation of 14 grams

What is the mass of the released silver, which is formed during the oxidation of 14 grams of butanal with silver oxide weighing 64 grams?

Let’s write the reaction equation:
CH3-CH2-CH2-COH + Ag2O = CH3-CH2-CH2-COOH + 2Ag.
We have given a number of two initial substances, which means that one of them is in excess, let’s determine which one:
From the reaction equation:
ν (С4Н8О) = ν (Ag2O).
m (С4Н8О) / M (С4Н8О) = m (Ag2O) / M (Ag2O).
Let’s determine how much silver oxide is needed for the oxidation of butanal:
m (Ag2O) = m (C4H8O) * M (Ag2O) / M (C4H8O).
Molar masses:
M (Ag) = 108 g / mol.
M (Ag2O) = 108 * 2 + 16 = 232 g / mol.
M (C4H8O) = 12 * 4 + 1 * 8 + 16 = 72 g / mol.
m (Ag2O) = 14 * 232/72 = 45.1 g, which means we have an excess of silver oxide, further calculations will be carried out using butanal.
From the reaction equation:
ν (С4Н8О) = ν (Ag) / 2.
m (C4H8O) / M (C4H8O) = m (Ag) / (2 * M (Ag)).
m (Ag) = m (C4H8O) * 2 * M (Ag) / M (C4H8O).
m (Ag) = 14 * 2 * 108/72 = 42 g.
Answer: 42 g of silver will stand out.