Em tham khảo nha :
\(\begin{array}{l}
18)\\
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{Fe}} = \dfrac{{11,2}}{{56}} = 0,2mol\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,2mol\\
{m_{FeC{l_2}}} = 0,2 \times 127 = 25,4g\\
b)\\
{n_{{H_2}}} = {n_{Fe}} = 0,2mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
c)\\
{n_{HCl}} = 2{n_{Fe}} = 0,4mol\\
{C_{{M_{HCl}}}} = \dfrac{{0,4}}{{0,2}} = 2M\\
19)\\
F{e_3}{O_4} + 4{H_2} \to 3Fe + 4{H_2}O\\
CuO + {H_2} \to Cu + {H_2}O\\
hh:F{e_3}{O_4}(a\,mol),CuO(b\,mol)\\
\left\{ \begin{array}{l}
232a + 80b = 54,4\\
3 \times 56a + 64b = 40
\end{array} \right.\\
\Rightarrow a = 0,2;b = 0,1\\
{n_{{H_2}}} = 4{n_{F{e_3}{O_4}}} + {n_{CuO}} = 0,9mol\\
{V_{{H_2}}} = 0,9 \times 22,4 = 20,16l
\end{array}\)
