in 100g oleum......20gSO3......80gH2SO4
200g.......................x=40g...........y=160g
dar, SO3 existent, produce o masa suplimentara de acid, conform reactiei date:
80gSO3..........98gH2SO4
40g............................x= 49g acid
m,total acid= 160+49=> 209g--acid
concentratia oleumului devine
c= mdx100/ms= 209x100/200= 104,5%
a doua solutie, trebuie sa contina 98g acid in 1oogsolutie
100g oleum...........104,5g acid
x..................................98g
x= 93,8g.....
pana la ms= 100g, trebuie sa se aduge peste 100g oleum, apa:
m,apa= 100g sol-93,8g oleum=6,2g