a,
Gọi $x$, $y$ là số mol $Mg$, $MgO$
$\Rightarrow 24x+40y=14$ $(1)$
$Mg+H_2SO_4\to MgSO_4+H_2$
$MgO+H_2SO_4\to MgSO_4+H_2O$
$n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)$
$\Rightarrow x=0,25$ $(2)$
$(1)(2)\Rightarrow x=0,25; y=0,2$
$m_{Mg}=0,25.24=6g$
$m_{MgO}=0,2.40=8g$
b,
$n_{H_2SO_4}=x+y=0,45(mol)$
$\Rightarrow m_{dd H_2SO_4}=0,45.98:4,9\%=900g$
$\Rightarrow m_{dd\text{spu}}=14+900-0,25.2=913,5g$
$n_{MgSO_4}=0,45(mol)$
$\Rightarrow C\%_{MgSO_4}=\dfrac{0,45.120.100}{913,5}=5,9\%$
