Theo giả thiết ta có : \(\left\{{}\begin{matrix}nAl2O3=\dfrac{20,4}{102}=0,2\left(mol\right)\\nMgO=\dfrac{8}{40}=0,2\left(mol\right)\\nNaOH=0,4.0,5=0,2\left(mol\right)\end{matrix}\right.\)
PTHH : \(\left\{{}\begin{matrix}Al2O3+3H2SO4->Al2\left(SO4\right)3+3H2O\left(1\right)\\MgO+H2SO4->MgSO4+H2O\left(2\right)\\2NaOH+H2SO4->Na2SO4+2H2O\left(3\right)\end{matrix}\right.\)
Theo 3 PTHH ta có : \(\left\{{}\begin{matrix}nH2SO4\left(1\right)=3nAl2O3=0,6\left(mol\right)\\nH2SO4\left(2\right)=nMgO=0,2\left(mol\right)\\nH2SO4\left(3\right)=\dfrac{1}{2}nNaOH=0,1\left(mol\right)\end{matrix}\right.\)
=> C%ddH2SO4(bđ) = \(\dfrac{\left(0,6+0,2+0,1\right).98}{122,5}.100\%=72\%\)
