Đáp án:

\({m_{{P_2}{O_5}}} = 2,84{\text{ gam}}\)

\({V_{{O_2}}} = 1,12{\text{ lít}}\)

\({m_{KMn{O_4}}}= 17,538{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(4P + 5{{\text{O}}_2}\xrightarrow{{{t^o}}}2{P_2}{O_5}\)

Ta có:

\({n_P} = \frac{{1,24}}{{31}} = 0,04{\text{ mol}} \to {{\text{n}}_{{P_2}{O_5}}} = \frac{1}{2}{n_P} = 0,02{\text{ mol}}\)

\({n_{{O_2}}} = \frac{5}{4}{n_P} = \frac{5}{4}.0,04 = 0,05{\text{ mol}}\)

\( \to {m_{{P_2}{O_5}}} = 0,02.(31.2 + 16.5) = 2,84{\text{ gam}}\)

\({V_{{O_2}}} = 0,05.22,4 = 1,12{\text{ lít}}\)

\(2KMn{O_4}\xrightarrow{{{t^o}}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)

\( \to {n_{KMn{O_4}{\text{ lt}}}} = 2{n_{{O_2}}} = 0,1{\text{ mol}}\)

Vì hao hụt \(10\%\) nên

\({n_{KMn{O_4}}} = \frac{{0,1}}{{100\%  - 10\% }} = 0,111{\text{ mol}}\)

\( \to {m_{KMn{O_4}}} = 0,111.(39 + 55 + 16.4) = 17,538{\text{ gam}}\)