Đáp án:

1,

a, \({V_{{H_2}}} = 5,6l\)

b, \({m_{Fe}} = 5,6g\) và \({m_{Cu}} = 6,4g\)

2,

a, \({V_{{H_2}}} = 8,96l\)

b, \({m_{HCl}} = 29,2g\)

c, \(C\% MgC{l_2} = \)4,59% và \(C\% AlC{l_3} = \)12,9%

Giải thích các bước giải:

1,

\(\begin{array}{l}
CuO + {H_2} \to Cu + {H_2}O\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
{m_{CuO}} = 16 \times \dfrac{{50}}{{100}} = 8g\\
 \to {n_{CuO}} = 0,1mol\\
 \to {m_{F{e_2}{O_3}}} = 8g \to {n_{F{e_2}{O_3}}} = 0,05mol\\
 \to {n_{{H_2}}} = {n_{CuO}} + 3{n_{F{e_2}{O_3}}} = 0,25mol\\
 \to {V_{{H_2}}} = 5,6l\\
{n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,1mol\\
 \to {m_{Fe}} = 5,6g\\
{n_{Cu}} = {n_{CuO}} = 0,1mol\\
 \to {m_{Cu}} = 6,4g
\end{array}\)

2,

\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{m_{Mg}} = \dfrac{4}{{13}} \times 7,8 = 2,4g\\
 \to {n_{Mg}} = 0,1mol\\
 \to {m_{Al}} = 5,4g \to {n_{Al}} = 0,2mol\\
 \to {n_{{H_2}}} = {n_{Mg}} + 1,5{n_{Al}} = 0,4mol\\
 \to {V_{{H_2}}} = 8,96l\\
{n_{HCl}} = 2{n_{Mg}} + 3{n_{Al}} = 0,8mol\\
 \to {m_{HCl}} = 29,2g\\
{m_{{\rm{HCldd}}}} = \dfrac{{29,2 \times 100}}{{14,6}} = 200g\\
 \to {m_{{\rm{dd}}}} = 7,8 + 200 - 0,4 \times 2 = 207g\\
 \to C\% MgC{l_2} = \dfrac{{0,1 \times 95}}{{207}} \times 100\%  = 4,59\% \\
 \to C\% AlC{l_3} = \dfrac{{0,2 \times 133,5}}{{207}} \times 100\%  = 12,9\% 
\end{array}\)