Đáp án:

\({m_{C{H_4}}} = 1,6{\text{ gam;}}{{\text{m}}_{{C_2}{H_4}}} = 0,7{\text{ gam}}\)

\(\% {m_{C{H_4}}} = 69,56\% ;\% {m_{{C_2}{H_4}}} = 30,44\% \)

\({V_{kk}} = 30,8{\text{ lít}}\)

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:

\({n_{B{{\text{r}}_2}}} = \frac{4}{{80.2}} = 0,025{\text{ mol = }}{{\text{n}}_{{C_2}{H_5}}}\)

\({n_{hh}} = \frac{{2,8}}{{22,4}} = 0,125{\text{ mol}} \to {{\text{n}}_{C{H_4}}} = 0,125 - 0,025 = 0,1{\text{ mol}}\)

\( \to {m_{C{H_4}}} = 0,1.16 = 1,6{\text{ gam;}}{{\text{m}}_{{C_2}{H_4}}} = 0,025.28 = 0,7{\text{ gam}}\)

\(\% {m_{C{H_4}}} = \frac{{1,6}}{{1,6 + 0,7}} = 69,56\%  \to \% {m_{{C_2}{H_4}}} = 30,44\% \)

Đốt cháy hỗn hợp

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

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

\( \to {n_{{O_2}}} = 3{n_{{C_2}{H_4}}} + 2{n_{C{H_4}}} = 0,025.3 + 0,1.2 = 0,275{\text{ mol}}\)

\( \to {V_{{O_2}}} = 0,275.22,4 = 6,16{\text{ lít}}\)

\( \to {V_{kk}} = 5{V_{{O_2}}} = 6,16.5 = 30,8{\text{ lít}}\)

`n_{Br_2}=(4)/(160)=0,025(mol)`

PTHH:

`C_2H_4+Br_2->C_2H_4Br_2`

`0,025←0,025(mol)`

(`CH_4` không phản ứng)

`a,`

Theo PTHH : `n_{C_2H_4}=0,025(mol)`

`=>m_{C_2H_4}=0,025.28=0,7(g)`

Ta có : `n_{CH_4}+n_{C_2H_4}=(2,8)/(22,4)=0,125`

`=>n_{CH_4}=0,125-0,025=0,1(mol)`

`=>m_{CH_4}=0,1.16=1,6(g)`

Phần trăm thể tích cũng là phần trăm số mol

`\%V_{C_2H_4}=\%n_{C_2H_4}=(0,025)/(0,125).100\%=20\%`

`=>\%V_{CH_4}=(100-20)\%=80\%`

`b,`

PTHH:

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

`0,025->0,075(mol)`

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

`0,1->0,2(mol)`

`=>n_{O_2}=0,2+0,075=0,275(mol)`

`=>V_{O_2}=0,275.22,4=6,16(lít)`

`=>V_{Kk}=6,16.5=30,8(lít)`