n1 moli CH3-Cl + 2NH3 ⇒ CH3-NH2 + NH4Cl
n2 moli 2CH3-Cl + 2NH3 ⇒(CH3)2NH + 2NH4Cl
n3 moli 3CH3-Cl + 2NH3 ⇒(CH3)3N + 3NH4Cl
n 1/1 = n2/(2·5) = n3/(3·2) = n/17 n1 = n/17 n2 = 10n/17 n3 = 6n/17
31n/17 + 45n/17 + 59n/17 = 374 135n/17 = 374 n = 47 moli
nNH3 pur = 2n1 + n2 + 2/3n3 = 2·2,77+ 10·2,77 + 6·2,77 = ≈50moli
m = 50·17 = 850g
m brut = 850·100/90 = 55,4 g
