PTHH: \(MgO+2HNO_3\rightarrow Mg\left(NO_3\right)_2+H_2O\) (1)
\(3Mg+8HNO_3\rightarrow3Mg\left(NO_3\right)_2+2NO+4H_2O\) (2)
Ta có: \(n_{NO}=\dfrac{0,336}{22,4}=0,015\left(mol\right)\) \(\Rightarrow n_{Mg}=0,0225\left(mol\right)\)
\(\Rightarrow\%m_{Mg}=\dfrac{0,0225\cdot24}{9,4}\cdot100\%\approx5,74\%\) \(\Rightarrow\%m_{MgO}=94,26\%\)
Theo PTHH: \(\left\{{}\begin{matrix}n_{Mg\left(NO_3\right)_2\left(2\right)}=0,0225mol\\n_{Mg\left(NO_3\right)_2\left(1\right)}=n_{MgO}=\dfrac{9,4-m_{Mg}}{40}=0,2215\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\Sigma n_{Mg\left(NO_3\right)_2}=0,244\left(mol\right)\) \(\Rightarrow m_{Mg\left(NO_3\right)_2}=0,244\cdot148=36,112\left(g\right)\)