Đáp án:
Bạn tham khảo lời giải ở dưới nhé !!!
Giải thích các bước giải:
\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{HCl}} = 0,08mol\\
1.\\
\to {n_{Fe}} = \dfrac{1}{2}{n_{HCl}} = 0,04mol\\
\to {m_{Fe}} = 2,24g\\
\to {m_C} = {m_{Cu}} = 8,24g\\
\to \% {m_{Fe}} = \dfrac{{2,24}}{{10,48}} \times 100\% = 21,37\% \\
\to \% {m_{Cu}} = 100\% - 21,37\% = 78,63\% \\
2.\\
Cu + 2{H_2}S{O_4} \to {\rm{CuS}}{O_4} + S{O_2} + 2{H_2}O\\
{\rm{CuS}}{O_4} + 2NaOH \to Cu{(OH)_2} + N{a_2}S{O_4}\\
Cu{(OH)_2} \to CuO + {H_2}O\\
{n_{{H_2}}} = \dfrac{1}{2}{n_{HCl}} = 0,04mol \to {V_{{H_2}}} = 0,896l\\
{n_{S{O_2}}} = 0,135mol\\
{n_{Cu}} = 0,129mol\\
\to {n_{S{O_2}}} > {n_{Cu}} \to {n_{S{O_2}}}dư\\
\to {n_{{\rm{CuS}}{O_4}}} = {n_{Cu}} = 0,129mol\\
\to {n_{Cu{{(OH)}_2}}} = {n_{{\rm{CuS}}{O_4}}} = 0,129mol\\
\to {n_{CuO}} = {n_{Cu{{(OH)}_2}}} = 0,129mol\\
\to {m_F} = {m_{CuO}} = 10,32g
\end{array}\)
