Em tham khảo nha :
\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}(1)\\
Fe + 2HCl \to FeC{l_2} + {H_2}(2)\\
{n_{HCl}} = 0,5 \times 1 = 0,5mol\\
{n_{{H_2}}} = \dfrac{{{n_{HCl}}}}{2} = 0,25mol\\
{V_{{H_2}}} = 0,25 \times 22,4 = 5,6l\\
b)\\
{n_{Fe}} = \dfrac{{11,2}}{{56}} = 0,2mol\\
{n_{HCl(2)}} = 2{n_{Fe}} = 0,4mol\\
{n_{HCl(1)}} = 0,5 - 0,4 = 0,1mol\\
{n_{Zn}} = \dfrac{{{n_{HCl(1)}}}}{2} = 0,05mol\\
{m_{Zn}} = 0,05 \times 65 = 3,25g\\
c)\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,05mol\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,2mol\\
{C_{{M_{ZnC{l_2}}}}} = \dfrac{{0,05}}{{0,5}} = 0,1M\\
{C_{{M_{FeC{l_2}}}}} = \dfrac{{0,2}}{{0,5}} = 0,4M
\end{array}\)
