Đáp án:
a) \({m_{{H_2}O}} = 5,4{\text{ gam}}\)
b) \(\% {m_{Fe}} = 30,29\% ;\% {m_{F{e_3}{O_4}}} = 69,71\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(F{e_3}{O_4} + 4{H_2}\xrightarrow{{{t^o}}}3Fe + 4{H_2}O\)
Ta có:
\({n_{{H_2}}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol;}}{{\text{n}}_{F{e_3}{O_4}}} = \frac{{46,4}}{{56.3 + 16.4}} = 0,2{\text{ mol > }}\frac{1}{4}{n_{{H_2}}}\)
Vậy oxit sắt dư
\( \to {n_{{H_2}}} = {n_{{H_2}O}} = 0,3{\text{ mol}} \to {m_{{H_2}O}} = 0,3.18 = 5,4{\text{ gam}}\)
Ta có:
\({n_{Fe}} = \frac{3}{4}{n_{{H_2}}} = 0,225{\text{ mol;}}{{\text{n}}_{F{e_3}{O_4}{\text{ dư}}}} = 0,2 - \frac{{0,3}}{4} = 0,125{\text{ mol}}\)
\( \to {m_{Fe}} = 0,225.56 = 12,6{\text{ gam; }}{{\text{m}}_{F{e_3}{O_4}}} = 0,125.232 = 29{\text{ gam}}\)
\( \to \% {m_{Fe}} = \frac{{12,6}}{{12,6 + 29}} = 30,29\% \to \% {m_{F{e_3}{O_4}}} = 69,71\% \)
