1.
n(hh)=0,25 mol
2H2S+3O2--->2SO2+2H2O
x-------->1,5x---->x
do oxi dư =>0,25-x>1,5x => x<0,1 mol
có nSO2=x
SO2+Br2+2H2O--->H2SO4+2HBr
có nSO2=nBr2=0,05 (nhận)
=> nSO2=0,05
=> nH2S=0,05 => nO2=0,2
2.
nH2SO4.2SO3=[tex]\frac{200}{258}[/tex]
nH2SO4.3SO3=[tex]\frac{100}{338}[/tex]
=> nSO3=[tex]2\frac{200}{258}+3\frac{100}{338}[/tex]
=> %mSO3=80.nSO3/300~65%