Giải thích các bước giải:

\(\begin{array}{l}
3.\\
a)Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
b)\\
{n_{{H_2}}} = 0,2mol\\
 \to {n_{Fe}} = {n_{{H_2}}} = 0,2mol\\
 \to {m_{Fe}} = 0,2 \times 56 = 11,2g\\
 \to {m_{Cu}} = 24 - 11,2 = 12,8g\\
 \to \% {m_{Fe}} = \dfrac{{11,2}}{{24}} \times 100\%  = 46,67\% \\
 \to \% {m_{Cu}} = \dfrac{{12,8}}{{24}} \times 100\%  = 53,33\% \\
c)\\
{n_{{H_2}S{O_4}}} = {n_{Fe}} = 0,2mol\\
 \to {m_{{H_2}S{O_4}}} = 19,6g\\
 \to {m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{19,6}}{{19,6\% }} \times 100\%  = 100g
\end{array}\)

Cu không tác dụng với H2SO4 ở nhiệt độ thường

\(\begin{array}{l}
4.\\
a)2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
b)\\
{n_{{H_2}}} = 0,3mol\\
 \to {n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol\\
 \to {m_{Al}} = 0,2 \times 27 = 5,4g\\
 \to {m_{Ag}} = 16,2 - 5,4 = 10,8g\\
 \to \% {m_{Al}} = \dfrac{{5,4}}{{16,2}} \times 100\%  = 33,3\% \\
 \to \% {m_{Ag}} = \dfrac{{10,8}}{{16,2}} \times 100\%  = 66,7\% \\
c)\\
{n_{{H_2}S{O_4}}} = \dfrac{3}{2}{n_{Al}} = 0,3mol\\
 \to {m_{{H_2}S{O_4}}} = 29,4g\\
 \to {m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{29,4}}{{7,53\% }} \times 100\%  = 390g
\end{array}\)

Bạc không tác dụng với H2SO4 ở nhiệt độ thường

\(\begin{array}{l}
7.\\
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
CuO + {H_2}S{O_4} \to C{\rm{uS}}{O_4} + {H_2}O\\
{n_{{H_2}}} = 0,6mol\\
 \to {n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,4mol\\
 \to {m_{Al}} = 10,8g\\
 \to {m_{CuO}} = 8g \to {n_{CuO}} = 0,1mol\\
 \to \% {m_{Al}} = \dfrac{{10,8}}{{18,8}} \times 100\%  = 57,45\% \\
 \to \% {m_{CuO}} = \dfrac{8}{{18,8}} \times 100\%  = 42,55\% \\
b)\\
{n_{{H_2}S{O_4}}} = \dfrac{3}{2}{n_{Al}} + {n_{CuO}} = 0,7mol\\
 \to {m_{{H_2}S{O_4}}} = 68,6g\\
 \to {m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{68,6}}{{19,6\% }} \times 100\%  = 350g\\
c)\\
{n_{{\rm{CuS}}{{\rm{O}}_4}}} = {n_{CuO}} = 0,1mol \to {m_{{\rm{CuS}}{{\rm{O}}_4}}} = 16g\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{1}{2}{n_{Al}} = 0,2mol \to {m_{A{l_2}{{(S{O_4})}_3}}} = 68,4g\\
 \to {m_{{\rm{dd}}muối}} = {m_{hỗnhợp}} + {m_{{\rm{dd}}{H_2}S{O_4}}} - {m_{{H_2}}} = 367,6g\\
 \to C{\% _{{\rm{CuS}}{{\rm{O}}_4}}} = \dfrac{{16}}{{367,6}} \times 100\%  = 4,35\% \\
 \to C{\% _{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{68,4}}{{367,6}} \times 100\%  = 18,6\% \\
\end{array}\)