Câu 4 :
a. \(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{100.9,8\%}{98}=0,1\left(mol\right)\)
PTHH : MgO + H2SO4 -> MgSO4 + H2O
0,05 0,05
Ta thấy : \(\dfrac{0.05}{1}< \dfrac{0.1}{1}\) => MgO đủ , H2SO4 dư
b. \(C\%_{dd_{MgSO_4}}=\dfrac{0,05.120}{2+100}.100=5,88\%\)