PTHH: \(2NaOH+MgSO_4\rightarrow Mg\left(OH\right)_2\downarrow+Na_2SO_4\)
\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)
Ta có: \(n_{NaOH}=0,2\cdot3=0,6\left(mol\right)\) \(\Rightarrow n_{Mg\left(OH\right)_2}=0,3mol=n_{MgO}\)
\(\Rightarrow m_{MgO}=0,3\cdot40=12\left(g\right)\)