calculati masa de amestec nitrant , ce contine 63% HNO3, necesar nitrarii a 312g benzen



Răspuns :

n = m / M = 312 / 78 = 4 moli benzen
Daca este mononitrare => 4 moli HNO3 react. => m = 4*63 = 252g HNO3
           63x / 100 = 252 => x = 400g amestec nitrant.