Đáp án:
\({V_{{O_2}{\text{ dư}}}} = 2,8{\text{ lít}}\)
\({m_{{P_2}{O_5}}} = 14,2{\text{ gam}}\)
\({m_{KMn{O_4}}} = 39,5{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(4P + 5{O_2}\xrightarrow{{{t^o}}}2{P_2}{O_5}\)
Ta có:
\({n_P} = \frac{{6,2}}{{31}} = 0,2{\text{ mol < }}\frac{4}{5}{n_{{O_2}}}\)
Do vậy \(O_2\) dư.
\( \to {n_{{O_2}{\text{ dư}}}} = 0,375 - \frac{5}{4}.0,2 = 0,125{\text{ mol}}\)
\( \to {V_{{O_2}{\text{ dư}}}} = 0,125.22,4 = 2,8{\text{ lít}}\)
\({n_{{P_2}{O_5}}} = \frac{1}{2}{n_P} = 0,1{\text{ mol}}\)
\( \to {m_{{P_2}{O_5}}} = 0,1.(31.2 + 16.5) = 14,2{\text{ gam}}\)
\(2KMn{O_4}\xrightarrow{{{t^o}}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)
\( \to {n_{KMn{O_4}}} = 2{n_{{O_2}}} = 0,25{\text{ mol}}\)
\( \to {m_{KMn{O_4}}} = 0,25.(39 + 55 + 16.4) = 39,5{\text{ gam}}\)
