Giải:
Mình gộp 2 câu với nhau làm cho dễ nhé
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*Hòa tan oleum vào H2O:
H2SO4.nSO3+H2O->(n+1)H2SO4(1)
0,16/n+1..................................0,16........(mol)
*Trung hòa oleum bằng dd KOH:
n(KOH)=0,08.4=0,32(mol)
2KOH+H2SO4->K2SO4+2H2O(2)
0,32..........0,16....................................(mol)
Theo PTHH:n(H2SO4)=0,16(mol)
Theo PTHH(1):n(H2SO4.nSO3)=0,16/n+1=13,8/98+80n
=>n=2
=>CTPT oleum là:H2SO4.2SO3
=>%SO3(oleum)=80.2/(98+80.2).100%=62%