Gọi CT oleum là $H_2SO_4.nSO_3$ GS có a molMol $KOH$ =0,8.0,1=0,08 mol
$SO_3$ + $H_2O$ \rightarrow $H2SO_4$
an mol =>an mol
Tổng mol $H_2SO_4$=an+a
$H_2SO_4$ + 2$KOH$ \rightarrow $K_2SO_4$ + $H_2O$
0,04 mol<=0,08 mol
=>an+a=0,04
Mà m oleum=3,38
=>a(98+80n)=3,38
=>an=0,03;a=0,01=>n=3 CT oleum $H_2SO_4.3H_2O$
b) GS cần b mol oleum
=>tông mol $H2SO4$ =4b mol=>m$H_2SO_4$=392b
=>mddsau=392b+200g
=>C%=392b/(392b+200)=0,1=>b=0,05668 mol
=>m=19,16g