Đáp án:
\({{\text{m}}_P} = 24,8{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(4P + 5{O_2}\xrightarrow{{{t^o}}}2{P_2}{O_5}\)
Ta có: \({n_P}:{n_{{O_2}}} = 4:5\)
\({n_{{P_2}{O_5}}} = \frac{{56,8}}{{31.2 + 16.5}} = 0,4{\text{ mol}} \to {{\text{n}}_P} = 2{n_{{P_2}{O_5}}} = 0,8{\text{ mol}} \to {{\text{m}}_P} = 0,8.31 = 24,8{\text{ gam}}\)
\({n_{{O_2}}} = \frac{5}{4}{n_P} = 1{\text{ mol}} \to {{\text{V}}_{{O_2}}} = 1.22,4 = 22,4{\text{ lít}} \to {{\text{V}}_{kk}} = \frac{{{V_{{O_2}}}}}{{20\% }} = 112{\text{ lít}}\)
\(C{H_4} + 2{O_2}\xrightarrow{{{t^o}}}C{O_2} + 2{H_2}O\)
\({n_{C{H_4}}} = \frac{1}{2}{n_{{O_2}}} = 0,5{\text{ mol}} \to {V_{C{H_4}}} = 0,5.22,4 = 11,2{\text{ lít}}\)
Cách 2 tính khối lượng P
BTKL:
\({m_P} + {m_{{O_2}}} = {m_{{P_2}{O_5}}} \to {m_P} + 1.32 = 56,8 \to {m_P} = 56,8 - 32 = 24,8{\text{ gam}}\)
