Question

Prin nitrarea fenolului se formeaza un produs ce contine 15,21 % N.Masa de fenol de puritate 50% necesara obtinerii a 552 kg de produs de nitrare,daca randamentul este de 75%,este: d)752kg

Answer 1

C6H6O + nHNO3 -> C6H6-nO(NO2)n
M=12*6+6-n+16+46n=45n+94
45n+94...............100%
14n...................15,21%
45n+94=1400n/15,21=92n
(92-45)n=94 => 47n=94 => n=2
produsul: dinitrofenolul
75/100=552/mteoretica => mteoretica=552*100/75=736kg
94kg............................184kg
C6H6O + 2HNO3 -> C6H4O(NO2)2 + 2H2O
x=376kg.....................736kg
50/100=376/mimpura => mimpura=376*100/50=752kg fenol impur