Calculate the mass of silver that precipitates when a silver mirror reacts 4.4 g of ethanal

Calculate the mass of silver that precipitates when a silver mirror reacts 4.4 g of ethanal with an ammoniacal solution of silver oxide.

Let’s write down the solution:
Let’s compose the reaction equation:
СН3 – СОН + Ag2O = 2Ag + CH3COOH – qualitative reaction to aldehyde, ethanal is oxidized, silver and acetic acid are released;
Let’s make calculations using the formulas:
M (CH3COH) = 12 * 2 + 4 + 16 = 44 g / mol;
M (Ag) = 107.8 g / mol;
Let us calculate the number of moles of ethanal if its mass is known:
Y (CH3COH) = m (CH3COH) / M (CH3COH);
Y (CH3COH) = 4.4 / 44 = 0.1 mol;
Let’s make a proportion according to the reaction equation:
0.1 mol (CH3COH) – X mol (Ag);
-1 mol – 2 mol hence, X mol (Ag) = 0.1 * 2/1 = 0.2 mol;
Let’s determine the mass of silver:
Y (Ag) = m (Ag) / M (Ag); m (Ag) = 0.2 * 107.8 = 21.56 g.
Answer: When a silver mirror reacts, silver with a mass of 21.56 g is released.