$a)PTHH:Mg+2HCl\to MgCl_2+H_2$

$MgO+2HCl\to MgCl_2+H_2O$

$b)n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)$

Theo PT: $n_{Mg}=n_{H_2}=0,1(mol)$

$\Rightarrow m_{Mg}=0,1.24=2,4(g)$

$\Rightarrow \%m_{Mg}=\dfrac{2,4}{6,4}.100\%=37,5\%$

$\Rightarrow \%m_{MgO}=100-37,5=62,5\%$

$c)n_{MgO}=\dfrac{6,4-2,4}{40}=0,1(mol)$

Theo PT: $n_{HCl}=2n_{Mg}+2n_{MgO}=0,2+0,2=0,4(mol)$

$\Rightarrow m_{dd_{HCl}}=\dfrac{0,4.36,5}{20\%}=73(g)$

$\Rightarrow V_{dd_{HCl}}=\dfrac{73}{1,1}\approx 66,36(ml)$

Em tham khảo nha:

\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
b)\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,1\,mol\\
\% {m_{Mg}} = \dfrac{{0,1 \times 24}}{{6,4}} \times 100\%  = 37,5\% \\
\% {m_{MgO}} = 100 - 37,5 = 62,5\% \\
c)\\
{n_{MgO}} = \dfrac{{6,4 - 2,4}}{{40}} = 0,1\,mol\\
{n_{HCl}} = 0,1 \times 2 + 0,1 \times 2 = 0,4\,mol\\
{m_{{\rm{ddHCl}}}} = \dfrac{{0,4 \times 36,5}}{{20\% }} = 73g\\
{V_{{\rm{ddHCl}}}} = \dfrac{{73}}{{1,1}} = 66,36\,ml
\end{array}\)