nNa2O=0,25(mol)
pt: Na2O + H2SO4-> Na2SO4 + H2O
=> nH2SO4=0,25(mol)
=> mH2SO4=0,25.98=24,5(g)
\(\Rightarrow a=C\%=\dfrac{24,5.100}{200}=12,25\left(\%\right)\)
PTHH: Na2O + H2SO4 → Na2SO4 + H2O
\(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{Na_2O}=0,25\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,25\times98=24,5\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}=\dfrac{24,5}{200}\times100\%=12,25\%\)
Vậy a% = 12,25%