\(n_{SO_3}=\dfrac{9,6}{80}=0,12\left(mol\right)\) ; \(n_{H_2O}=\dfrac{90,4}{18}\approx5,02\left(mol\right)\)
PTHH : SO3 + H2O ---> H2SO4
Vì \(n_{H2O}>n_{SO3}\) => H2O dư
Theo PTHH : nH2SO4 = nSO3 = 0,12 (mol)
=> mH2SO4 = 0,12.98 = 11,76 (g)
Theo ĐLBTKL : mSO3 + mH2O = mH2SO4
=> mH2SO4 = 9,6 + 90,4 = 100 (g)
=> \(C\%_{ddH2SO4}=\dfrac{11,76}{100}.100\%=11,76\%\)