Đáp án:
\( {V_{{C_2}{H_4}{\text{ dư}}}} = 10,45333{\text{ lít}}\)
\( \ {m_{C{O_2}}} = 11,7333{\text{ gam;}}{{\text{m}}_{{H_2}O}} = 4,8{\text{ gam}}\)
\(m{_{{C_2}{H_4}}} = 3,7333{\text{ gam}}\)
Giải thích các bước giải:
\({C_2}{H_4} + 3{O_2}\xrightarrow{{{t^o}}}2C{O_2} + 2{H_2}O\)
Ta có:
\({n_{{C_2}{H_4}}} = \frac{{13,44}}{{22,4}} = 0,6{\text{ mol}}\)
\({n_{{O_2}}} = \frac{{8,96}}{{22,4}} = 0,4{\text{ mol < 3}}{{\text{n}}_{{C_2}{H_4}}}\)
Vậy \(C_2H_4\) dư
\({n_{{C_2}{H_4}{\text{ phản ứng}}}} = \frac{{{n_{{O_2}}}}}{3} = \frac{{0,4}}{3}\)
\( \to {n_{{C_2}{H_4}{\text{ dư}}}} = 0,6 - \frac{{0,4}}{3} = \frac{{1,4}}{3}\)
\( \to {V_{{C_2}{H_4}{\text{ dư}}}} = \frac{{1,4}}{3}.22,4 = 10,45333{\text{ lít}}\)
\({n_{C{O_2}}} = {n_{{H_2}O}} = \frac{{0,4}}{3}.2 = \frac{{0,8}}{3}\)
\( \to {m_{C{O_2}}} = \frac{{0,8}}{3}.44 = 11,7333{\text{ gam;}}{{\text{m}}_{{H_2}O}} = \frac{{0,8}}{3}.18 = 4,8{\text{ gam}}\)
\(m{_{{C_2}{H_4}}} = \frac{{0,4}}{3}.28 = 3,7333{\text{ gam}}\)
