Đáp án:
\({m_{C{H_4}}} = 3,02144{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(C{H_4} + 2{O_2}\xrightarrow{{{t^o}}}C{O_2} + 2{H_2}O\)
Ta có:
\({n_{{O_2}}} = \frac{{8,46}}{{22,4}} = 0,37768{\text{ mol}}\)
\( \to {n_{C{H_4}}} = \frac{1}{2}{n_{{O_2}}} = 0,18884{\text{ mol }}\)
\( \to {m_{C{H_4}}} = 0,18884.16 = 3,02144{\text{ gam}}\)
\({V_{C{O_2}}} = \frac{1}{2}{V_{{O_2}}} = 4,23{\text{ lít}}\)
\({n_{{H_2}O}} = {n_{{O_2}}} = 0,37768{\text{ mol}}\)
\( \to {m_{{H_2}O}} = 0,37768.18 = 6,79824{\text{ gam}}\)