Đáp án:
\( {m_P} = 1,86{\text{ gam}}\)
\( {V_{{O_2}}} = 1,68{\text{ lít}}\)
\( {m_{KCl{O_3}}} = 6,125{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(4P + 5{O_2}\xrightarrow{{{t^o}}}2{P_2}{O_5}\)
Ta có:
\({n_{{P_2}{O_5}}} = \frac{{4,26}}{{31.2 + 16.5}} = 0,03{\text{ mol}}\)
\( \to {n_P} = 2{n_{{P_2}{O_5}}} = 0,03.2 = 0,06{\text{ mol}}\)
\( \to {m_P} = 0,06.31 = 1,86{\text{ gam}}\)
\({n_{{O_2}}} = \frac{5}{2}{n_{{P_2}{O_5}}} = \frac{5}{2}.0,03 = 0,075{\text{ mol}}\)
\( \to {V_{{O_2}}} = 0,075.22,4 = 1,68{\text{ lít}}\)
Phản ứng xảy ra:
\(2KCl{O_3}\xrightarrow{{{t^o}}}2KCl + 3{O_2}\)
\( \to {n_{KCl{O_3}}} = \frac{2}{3}{n_{{O_2}}} = \frac{2}{3}.0,075 = 0,05{\text{ mol}}\)
\( \to {m_{KCl{O_3}}} = 0,05.(39 + 35,5 + 16.3) = 6,125{\text{ gam}}\)