PTHH: \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
Ta có: \(n_{MgO}=\dfrac{12}{40}=0,3\left(mol\right)=n_{H_2SO_4}=n_{MgSO_4}\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{H_2SO_4}=\dfrac{0,3\cdot98}{80}\cdot100\%=36,75\%\\C\%_{MgSO_4}=\dfrac{0,3\cdot120}{12+80}\cdot100\%\approx39,13\%\end{matrix}\right.\)