4)
Phản ứng xảy ra:
\(4P + 5{O_2}\xrightarrow{{{t^o}}}2{P_2}{O_5}\)
Ta có:
\({n_P} = \frac{{6,2}}{{31}} = 0,2{\text{ mol;}}{{\text{n}}_{{O_2}}} = \frac{{8,96}}{{22,4}} = 0,4{\text{ mol > }}\frac{5}{4}{n_P}\)
Vậy \(O_2\) dư, hiệu suất tính theo \(P\)
\( \to {n_{P{\text{ phản ứng}}}} = 0,2.80\% = 0,16{\text{ mol}} \to {{\text{n}}_{P{\text{ dứ}}}} = 0,04{\text{ mol}}\)
\( \to {n_{{P_2}{O_5}}} = \frac{1}{2}{n_{P{\text{ phản ứng}}}} = 0,08{\text{ mol}}\)
\( \to {m_{rắn}} = {m_{{P_2}{O_5}}} + {m_P} = 0,08.142 + 0,04.31 = 12,6{\text{ gam}}\)
5)
Phản ứng xảy ra:
\(3Fe + 2{O_2}\xrightarrow{{{t^o}}}F{e_3}{O_4}\)
\(2Mg + {O_2}\xrightarrow{{{t^o}}}2MgO\)
Ta có:
\({n_{Mg}} = \frac{{0,48}}{{24}} = 0,02{\text{ mol;}}{{\text{n}}_{{O_2}}} = \frac{{0,672}}{{22,4}} = 0,03{\text{ mol}}\)
\( \to {n_{Fe}} = \frac{3}{2}.({n_{{O_2}}} - \frac{1}{2}{n_{Mg}}) = 0,03{\text{ mol}}\)
\( \to {m_{Fe}} = 0,03.56 = 1,68{\text{ gam}}\)
