Đáp án:
\( {V_{{O_2}}} = 3,2{\text{ }}{{\text{m}}^3}\)
\( {m_{F{e_3}{O_4}}} = 16,57{\text{ kg}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(3Fe + 2{O_2}\xrightarrow{{{t^o}}}F{e_3}{O_4}\)
Ta có:
\({n_{Fe}} = \frac{{12}}{{56}}\)
\( \to {n_{{O_2}}} = \frac{2}{3}{n_{Fe}} = \frac{1}{7} \to {V_{{O_2}}} = \frac{1}{7}.22,4 = 3,2{\text{ }}{{\text{m}}^3}\)
\({n_{F{e_3}{O_4}}} = \frac{1}{3}{n_{Fe}} = \frac{1}{3}.\frac{{12}}{{56}} = \frac{1}{{14}}\)
\( \to {m_{F{e_3}{O_4}}} = \frac{1}{{14}}.(56.3 + 16.4) = 16,57{\text{ kg}}\)