Đáp án:
\(\begin{array}{l}
b)\\
\% {V_{C{H_4}}} = 66,7\% \\
\% {V_{{C_2}{H_4}}} = 33,3\% \\
c)\\
{V_{kk}} = 39,2l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
b)\\
{V_{C{H_4}}} = 2,24l\\
{V_{{C_2}{H_4}}} = 3,36 - 2,24 = 1,12l\\
\% {V_{C{H_4}}} = \dfrac{{2,24}}{{3,36}} \times 100\% = 66,7\% \\
\% {V_{{C_2}{H_4}}} = 100 - 66,7 = 33,3\% \\
c)\\
C{H_4} + 2{O_2} \xrughtarrow{t^0} C{O_2} + 2{H_2}O\\
{C_2}{H_4} + 3{O_2} \xrightarrow{t^0} 2C{O_2} + 2{H_2}O\\
{n_{C{H_4}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_{{C_2}{H_4}}} = \dfrac{{1,12}}{{22,4}} = 0,05mol\\
{n_{{O_2}}} = 2{n_{C{H_4}}} + 3{n_{{C_2}{H_4}}} = 0,35mol\\
{V_{{O_2}}} = 0,35 \times 22,4 = 7,84l\\
{V_{kk}} = 5{V_{{O_2}}} = 7,84 \times 5 = 39,2l
\end{array}\)