Đáp án:
\( \% {m_{Fe}} = 63,6\% ;\% {m_{Cu}} = 36,4\% \)
\( C{\% _{FeS{O_4}}} = 7,4\% \)
Giải thích các bước giải:
\(Cu\) không tan trong \(H_2SO_4\) loãng.
Phản ứng xảy ra:
\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)
Ta có:
\({n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol = }}{{\text{n}}_{Fe}}\)
\( \to {m_{Fe}} = 0,1.56 = 5,6{\text{ gam}}\)
\( \to \% {m_{Fe}} = \frac{{5,6}}{{8,8}}.100\% = 63,6\% \to \% {m_{Cu}} = 36,4\% \)
\({n_{FeS{O_4}}} = {n_{Fe}} = 0,1{\text{ mol}}\)
\( \to {m_{FeS{O_4}}} = 0,1.(56 + 96) = 15,2{\text{ gam}}\)
Bảo toàn khối lượng:
\({m_{Fe}} + {m_{dd\;{{\text{H}}_2}S{O_4}}} = {m_{dd}} + {m_{{H_2}}}\)
\( \to 5,6 + 200 = {m_{dd}} + 0,1.2\)
\( \to {m_{dd}} = 205,4{\text{ gam}}\)
\( \to C{\% _{FeS{O_4}}} = \frac{{15,2}}{{205,4}}.100\% = 7,4\% \)