Em tham khảo nha :
\(\begin{array}{l}
a)\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
FeO + 2HCl \to FeC{l_2} + {H_2}O\\
{n_{HCl}} = 0,4 \times 0,45 = 0,18mol\\
hh:MgO(a\,mol),FeO(b\,mol)\\
\left\{ \begin{array}{l}
40a + 72b = 4,88\\
2a + 2b = 0,18
\end{array} \right.\\
\Rightarrow a = 0,05mol;b = 0,04mol\\
{m_{MgO}} = 0,05 \times 40 = 2g\\
{m_{FeO}} = 4,88 - 2 = 2,88g\\
b)\\
MgC{l_2} + 2NaOH \to Mg{(OH)_2} + 2NaCl\\
FeC{l_2} + 2NaOH \to Fe{(OH)_2} + 2NaCl\\
{n_{Mg{{(OH)}_2}}} = {n_{MgC{l_2}}} = {n_{MgO}} = 0,05mol\\
{n_{Fe{{(OH)}_2}}} = {n_{FeC{l_2}}} = {n_{FeO}} = 0,04mol\\
Fe{(OH)_2} \to FeO + {H_2}O\\
{n_{FeO}} = {n_{Fe{{(OH)}_2}}} = 0,04mol\\
{m_{FeO}} = 0,04 \times 72 = 2,88g
\end{array}\)