CxH2xO + H2 ⇒ CxH(2x+2)O
nX + nY = 112/22,4 = 5 moli
R-CH=O + 2[Ag(NH3)2]OH ⇒ R-COOH + 2Ag↓ + 4NH3 + H2O
n aldehida = nAg/2 = 2moli
nZ = nX = 2moli 2 = 176/M M = 88g/mol
CxH2xO2 14x + 32 = 88 14x = 56 x = 4
X = butanal ; Y = butanona
mA = 5·72 = 360g