Đáp án:

\(\% {V_{C{H_4}}} = \% {V_{{C_2}{H_4}}} = 50\% \)

\({V_{{O_2}}} = 11,2{\text{ lít}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\({C_2}{H_4} + B{{\text{r}}_2}\xrightarrow{{}}{C_2}{H_4}B{{\text{r}}_2}\)

Ta có:

\({m_{B{r_2}}} = 160.10\%  = 16{\text{ gam}} \to {{\text{n}}_{B{{\text{r}}_2}}} = \frac{{16}}{{80.2}} = 0,1{\text{ mol = }}{{\text{n}}_{{C_2}{H_4}}}\)

\( \to {V_{{C_2}{H_4}}} = 0,1.22,4 = 2,24{\text{ lít}} \to {{\text{V}}_{C{H_4}}} = 2,24{\text{ lít}}\)

\( \to \% {V_{C{H_4}}} = \% {V_{{C_2}{H_4}}} = 50\% \)

Đốt cháy hỗn hợp

\(C{H_4} + 2{{\text{O}}_2}\xrightarrow{{{t^o}}}C{O_2} + 2{H_2}O\)

\({C_2}{H_4} + 3{{\text{O}}_2}\xrightarrow{{{t^o}}}2C{O_2} + 2{H_2}O\)

\( \to {V_{{O_2}}} = 2{V_{C{H_4}}} + 3{V_{{C_2}{H_4}}} = 2,24.3 + 2,24.5 = 11,2{\text{ lít}}\)

Đáp án+ Giải thích các bước giải:

`a)`

`C_2H_4 + Br_2 -> C_2H_4Br_2`

`n_{Br_2}={160.10%}/{160.100%}=0,1(mol)`

Theo pt : `n_{C_2H_4}=n_{Br_2}=0,1(mol)`

`-> %V_{C_2H_4}={0,1.22,4.100%}/{4,48}=50%`

`-> %V_{CH_4}= 100%-50%=50%`

`b)`

`CH_4 + 2O_2` $\xrightarrow{t^o}$ `CO_2 + 2H_2O(1)`

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

`n_{CH_4}=V/{22,4}={4,48-0,1.22,4}/{22,4}=0,1(mol)`

Theo pt `(1)` và `(2)` : $\Sigma$`n_{O_2} = 2n_{CH_4}+3n_{C_2H_4} = 2.0,1 + 3.0,1=0,5(mol)`

`-> V_{O_2}=n.22,4=0,5.22,4=11,2(l)`

$\text{#Cyrus}$