\(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)
\(n_{H2SO4}=\dfrac{19,6\%.100}{100\%.98}=0,2\left(mol\right)\)
Pt : \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
Xét tỉ lệ : \(\dfrac{0,05}{1}< \dfrac{0,2}{1}\Rightarrow H_2SO_4dư\)
Theo pt : \(n_{MgO\left(pư\right)}=n_{MgSO4}=0,05\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgSO4}=\dfrac{0,05.120}{2+100}.100\%=5,88\%\\C\%_{ddH2SO4\left(dư\right)}=\dfrac{\left(0,2-0,05\right).98}{2+100}.100\%=14,41\%\end{matrix}\right.\)