M,etanal= 44g/mol
M,Cu2O= 144g
1mol...............................................................1mol
CH3CHO + 2Cu(OH)2---> CH3COOH + Cu2O + 2H2O
44g.....................................................................144g
440g.......................................................................x= 1440g-masa teoretica
75/100= m,practica/m,teor---> m,pr= 1440x75/100g=......................calculeaza !