2R-CHO + 2Cu(OH)2 => 2R-COOH + Cu2O + H2O
100% ..................................... miu.aldehida
(100+37,21)% ........................ miu.aldehida+16
=> 16+miu = 1,3721miu
=> miu = 43 g/mol
R-CH=O = CnH2nO
=> 14n+16 = 43 => n = 2
=> C2H4O = etanal....
cred ca datele nu sunt corect scrise.....