Đáp án:
\({m_{Fe}} = 2,52{\text{ gam}}\)
\({m_{{O_2}}} = 0,96{\text{ gam}}\)
\( {m_{KCl{O_3}}} = 2,45{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(3Fe + 2{O_2}\xrightarrow{{{t^o}}}F{e_3}{O_4}\)
Ta có:
\({n_{F{e_3}{O_4}}} = \frac{{3,48}}{{56.3 + 16.4}} = 0,015{\text{ mol}}\)
\( \to {n_{Fe}} = 3{n_{F{e_3}{O_4}}} = 0,045{\text{ mol}}\)
\({n_{{O_2}}} = 2{n_{F{e_3}{O_4}}} = 0,015.2 = 0,03{\text{ mol}}\)
\( \to {m_{Fe}} = 0,045.56 = 2,52{\text{ gam}}\)
\({m_{{O_2}}} = 0,03.32 = 0,96{\text{ gam}}\)
Điều chế \(O_2\)
\(2KCl{O_3}\xrightarrow{{{t^o}}}2KCl + 3{O_2}\)
\( \to {n_{KCl{O_3}}} = \frac{2}{3}{n_{{O_2}}} = 0,02{\text{ mol}}\)
\( \to {m_{KCl{O_3}}} = 0,02.(39 + 35,5 + 16.3) = 2,45{\text{ gam}}\)