Find the mass of silver obtained by the reaction of butyraldehyde weighing 20 g with silver oxide weighing 50 g

Let’s write the equation:
H3C – CH2 – CH2 – COH + Ag2O = 2Ag + H3C – CH2 – CH2 – COOH;
Let’s make the calculations:
M (C4H8O) = 72 g / mol; M (Ag2O) = 231.6 g / mol; M (Ag) = 107.8 g / mol;
Determine the amount of moles of aldehyde, silver oxide:
Y (C4H8O) = m / M = 20/72 = 0.27 mol;
Y (Ag2O) = m / M = 50 / 231.6 = 0.21 mol;
Let’s make the proportion:
0.21 mol (Ag2O) – X mol (Ag);
-1 mol -2 mol from here, X mol (Ag) = 0.21 * 2/1 = 0.43 mol;
Find the mass of silver:
m (Ag) = Y * M = 0.43 * 107.8 = 46.54 g.