n1 kmoli C6H5-NH2 + 2(CH2)2O ⇒ C6H5-N(CH2-CH2-OH)2
n2 kmoli (CH2)2O
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final : n1 kmoli produs + (n2 - 2n1)kmoli oxid de etena
M produs = 181g/moli
teoretic : mt produs = 1370 ·4/3 n1 = 10kmoli
n1/(n2 - 2n1 ) = 2/1 n1 = 2n2 - 4n1 n2 = 2,5n1 = 25 kmoli
Δn = 25 - 20 = 5 kmoli %exces = 5·100/25 = 20%