\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,05\(\leftarrow\) 0,05 (mol)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
Theo ptpu: \(n_{Mg}=0,05\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,05.24=1,2\left(g\right)\)
\(\Rightarrow\%Mg=\dfrac{1,2.100\%}{9,2}=13,04\%\)
\(\Rightarrow\%MgO=100\%-13,04\%=86,96\%\)