$Mg+2H_2SO_4 \rightarrow MgSO_4 + SO_2+2H_2O$
$MgO + H_2SO_4 \rightarrow MgSO_4 + H_2O$
$n_{SO2}=\dfrac{0,448}{22,4}=0,02(mol)$
Ta có: $n_{SO2}=n_{Mg}=0,02(mol)$
$n_{MgO}=0,01(mol)$
$m_{Mg}=0,02.24=0,48g$
$m_{MgO}=0,01.40=0,4g$
%$m_{Mg}=\dfrac{0,48}{0,4+0,48}.100$%$=54,54$%
%$m_{MgO}=100$%$-54,54$%$=45,56$%