Đáp án:

Bạn tham khảo nha ! 

Giải thích các bước giải:

\(\begin{array}{l}
1.\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = 0,2mol\\
{n_{{H_2}S{O_4}}} = 0,05mol\\
 \to \dfrac{{{n_{Al}}}}{2} > \dfrac{{{n_{{H_2}S{O_4}}}}}{3} \to {n_{Al}}dư
\end{array}\)

\(\begin{array}{l}
a)\\
{n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,05mol\\
 \to {V_{{H_2}}} = 1,12l\\
b)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{1}{3}{n_{{H_2}S{O_4}}} = 0,0167mol\\
 \to C{M_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{0,0167}}{{0,1}} = 0,167M
\end{array}\)

\(\begin{array}{l}
2.\\
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{Mg}} = 0,2mol
\end{array}\)

Khí thoát ra là khí hidro

\(\begin{array}{l}
b)\\
{n_{HCl}} = 2{n_{Mg}} = 0,4mol\\
 \to C{M_{HCl}} = \dfrac{{0,4}}{{0,4}} = 1M
\end{array}\)

\(\begin{array}{l}
c)\\
CuO + {H_2} \to Cu + {H_2}O\\
{n_{{H_2}}} = {n_{Mg}} = 0,2mol\\
{n_{CuO}} = 0,4mol\\
 \to {n_{CuO}} > {n_{{H_2}}} \to {n_{CuO}}dư\\
 \to {n_{Cu}} = {n_{CuO}} = {n_{{H_2}}} = 0,2mol\\
 \to {m_{Cu}} = 12,8g\\
 \to {n_{CuO(dư)}} = 0,2mol \to {m_{CuO(du)}} = 16g\\
 \to \% {m_{Cu}} = \dfrac{{12,8}}{{12,8 + 16}} \times 100\%  = 44,4\% \\
 \to \% {m_{CuO}} = \dfrac{{16}}{{12,8 + 16}} \times 100\%  = 55,6\% 
\end{array}\)

\(\begin{array}{l}
3.\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
a)\\
{n_{Zn}} = 0,2mol\\
 \to {n_{{H_2}}} = {n_{Zn}} = 0,2mol\\
 \to {V_{{H_2}}} = 4,48l\\
b)\\
{n_{HCl}} = 2{n_{Zn}} = 0,4mol\\
 \to {m_{HCl}} = 14,6g\\
 \to {m_{{\rm{dd}}HCl}} = \dfrac{{14,6}}{{14,6\% }} \times 100\%  = 100g
\end{array}\)

\(\begin{array}{l}
c)\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,2mol\\
 \to {m_{ZnC{l_2}}} = 27,2g\\
d)\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
{n_{F{e_2}{O_3}}} = 0,4mol\\
{n_{{H_2}}} = 0,2mol\\
 \to {n_{F{e_2}{O_3}}}dư\\
 \to {n_{Fe}} = \dfrac{2}{3}{n_{{H_2}}} = 0,13mol\\
 \to {m_{Fe}} = 7,28g
\end{array}\)