Đáp án:
\(\% {m_{Fe}} = 63,6\%; \% {m_{Cu}} = 36,4\% \)
\(V= 4,48{\text{ lít}}\)
\( {V_{dd\;{{\text{H}}_2}S{O_4}}} = 21,74{\text{ ml}}\)
Giải thích các bước giải:
\(Cu\) không tác dụng với \(HCl\)
Phản ứng xảy ra:
\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)
Ta có:
\({n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol = }}{{\text{n}}_{Fe}}\)
\( \to {m_{Fe}} = 0,1.56 = 5,6{\text{ gam}}\)
\( \to \% {m_{Fe}} = \frac{{5,6}}{{8,8}}.100\% = 63,6\% \to \% {m_{Cu}} = 36,4\% \)
\({m_{Cu}} = 8,8 - 5,6 = 3,2{\text{ gam}}\)
\( \to {n_{Cu}} = \frac{{3,2}}{{64}} = 0,05{\text{ mol}}\)
Cho hỗn hợp tác dụng với \(H_2SO_4\) đặc
\(2Fe + 6{H_2}S{O_4}\xrightarrow{{{t^o}}}F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\)
\(Cu + 2{H_2}S{O_4}\xrightarrow{{}}CuS{O_4} +SO_2+ 2{H_2}O\)
\( \to {n_{S{O_3}}} = \frac{3}{2}{n_{Fe}} + {n_{Cu}} = 0,1.\frac{3}{2} + 0,05 = 0,2{\text{ mol}}\)
\( \to V= {V_{S{O_2}}} = 0,2.22,4 = 4,48{\text{ lít}}\)
\({n_{{H_2}S{O_4}}} = 2{n_{S{O_2}}} = 0,4{\text{ mol}}\)
\( \to {m_{{H_2}S{O_4}}} = 0,4.98 = 39,2{\text{ gam}}\)
\( \to {m_{dd\;{{\text{H}}_2}S{O_4}}} = \frac{{39,2}}{{98\% }} = 40{\text{ gam}}\)
\( \to {V_{dd\;{{\text{H}}_2}S{O_4}}} = \frac{{40}}{{1,84}} = 21,74{\text{ ml}}\)
