Đáp án:
\({m_{{P_2}{O_5}}} = 7,1{\text{ gam}}\)
\({m_{S{O_2}}} = 19,2{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(4P + 5{O_2}\xrightarrow{{{t^o}}}2{P_2}{O_5}\)
\(S + {O_2}\xrightarrow{{{t^o}}}S{O_2}\)
Ta có:
\({n_P} = \frac{{3,1}}{{31}} = 0,1{\text{ mol;}}{{\text{n}}_S} = \frac{{9,6}}{{32}} = 0,3{\text{ mol;}}{{\text{n}}_{{O_2}}} = \frac{{13,44}}{{22,4}} = 0,6{\text{ mol}}\)
Vì \({n_{{O_2}}} > \frac{5}{4}{n_P} + {n_S}\) nên \(O_2\) dư.
\( \to {n_{{P_2}{O_5}}} = \frac{1}{2}{n_P} = 0,05{\text{ mol}}\)
\({n_{S{O_2}}} = {n_S} = 0,3{\text{ mol}}\)
\( \to {m_{{P_2}{O_5}}} = 0,05.(31.2 + 16.5) = 7,1{\text{ gam}}\)
\({m_{S{O_2}}} = 0,3.(32 + 16.2) = 19,2{\text{ gam}}\)
