Đáp án:

\(\begin{array}{l}
b)\\
\% {m_{Fe}} = 46,67\% \\
\% {m_{Cu}} = 53,33\% \\
c){m_{{H_2}S{O_4}}} = 9,8g\\
d)\\
{m_{Cu}} = 6,4g\\
{m_{CuO(dư)}} = 72g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
b)\\
{n_{{H_2}}} = 0,1mol\\
 \to {n_{Fe}} = {n_{{H_2}}} = 0,1mol\\
 \to {m_{Fe}} = 5,6g\\
 \to {m_{Cu}} = 6,4g\\
 \to \% {m_{Fe}} = \dfrac{{5,6}}{{12}} \times 100\%  = 46,67\% \\
 \to \% {m_{Cu}} = 100\%  - 46,67\%  = 53,33\% 
\end{array}\)

\(\begin{array}{l}
c)\\
{n_{{H_2}S{O_4}}} = {n_{Fe}} = 0,1mol\\
 \to {m_{{H_2}S{O_4}}} = 9,8g
\end{array}\)

\(\begin{array}{l}
d)\\
{H_2} + CuO \to Cu + {H_2}O\\
{n_{CuO}} = 1mol\\
 \to {n_{{H_2}}} < {n_{CuO}} \to {n_{CuO}}dư\\
 \to {n_{Cu}} = {n_{CuO}} = {n_{{H_2}}} = 0,1mol\\
 \to {m_{Cu}} = 6,4g\\
 \to {n_{CuO(dư)}} = 0,9mol\\
 \to {m_{CuO(dư)}} = 72g
\end{array}\)