Em tham khảo nha:

\(\begin{array}{l}
a)\\
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
b)\\
{n_{B{r_2}}} = \dfrac{{16}}{{160}} = 0,1\,mol\\
{n_{{C_2}{H_4}}} = {n_{B{r_2}}} = 0,1\,mol\\
{V_{{C_2}{H_4}}} = 0,1 \times 22,4 = 2,24l\\
{V_{C{H_4}}} = 11,2 - 2,24 = 8,96l\\
\% {V_{{C_2}{H_4}}} = \dfrac{{2,24}}{{11,2}} \times 100\%  = 20\% \\
\% {V_{C{H_4}}} = 100 - 20 = 80\% \\
c)\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
{C_2}{H_4} + 3{O_2} \xrightarrow{t^0} 2C{O_2} + 2{H_2}O\\
{n_{C{H_4}}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
{n_{{O_2}}} = 2{n_{C{H_4}}} + 3{n_{{C_2}{H_4}}} = 2 \times 0,4 + 0,1 \times 3 = 1,1\,mol\\
{V_{kk}} = 1,1 \times 5 \times 22,4 = 123,2l
\end{array}\)

Đáp án + Giải thích các bước giải:

$a\big)C_2H_4+Br_2\to C_2H_4Br_2$

$b\big)$

Theo PT: $n_{C_2H_4}=n_{Br_2}=\frac{16}{160}=0,1(mol)$

$\to \%V_{C_2H_4}=\frac{0,1.22,4}{11,2}.100\%=20\%$

$\to \%V_{CH_4}=100-20=80\%$

$c\big)$

$n_{CH_4}=\frac{11,2-22,4.0,1}{22,4}=0,4(mol)$

$CH_4+2O_2\xrightarrow{t^o}CO_2+2H_2O$

$C_2H_4+3O_2\xrightarrow{t^o}2CO_2+2H_2O$

Theo PT: $\sum n_{O_2}=2.0,4+3.0,1=1,1(mol)$

$\to V_{không\,khí}=\frac{1,1.22,4}{20\%}=123,2(l)$