Đáp án:
\(\% {m_{Fe}} = 83,8\% ; \% {m_{FeO}} = 16,2\% \)
\( m = 138{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2C{H_3}COOH + Fe\xrightarrow{{}}{(C{H_3}COO)_2}Fe + {H_2}\)
\(2C{H_3}COOH + FeO\xrightarrow{{}}{(C{H_3}COO)_2}Fe + {H_2}O\)
Ta có:
\({n_{{H_2}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol = }}{{\text{n}}_{Fe}}\)
\( \to {m_{Fe}} = 0,2.56 = 11,2{\text{ gam}} \to {{\text{m}}_{FeO}} = 13,36 - 11,2 = 2,16{\text{ gam}}\)
\( \to {n_{FeO}} = \frac{{2,16}}{{56 + 16}} = 0,03{\text{ mol}}\)
\(\% {m_{Fe}} = \frac{{11,2}}{{13,36}} = 83,8\% \to \% {m_{FeO}} = 16,2\% \)
\({n_{C{H_3}COOH}} = 2{n_{Fe}} + 2{n_{FeO}} = 0,2.2 + 0,03.2 = 0,46{\text{ mol}}\)
\( \to {m_{C{H_3}COOH}} = 0,46.60 = 27,6{\text{ gam}}\)
\( \to m = \frac{{27,6}}{{20\% }} = 138{\text{ gam}}\)