Đáp án + Giải thích các bước giải:

a.

Phản ứng xảy ra:

$C_2H_4+Br_2 \to C_2H_4Br_2$

b.

$n_{hh}=\dfrac{2,24}{22,4}=0,1(mol)$

$n_{C_2H_4}=n_{Br_2}= \dfrac{8}{160}=0,05(mol)$

$n_{CH_4}=0,1-0,05=0,05(mol)$

$\%V_{C_2H_4}=\%V_{CH_4}=\dfrac{0,05}{0,1}.100\%=50\%$

$CH_4+2O_2\xrightarrow{t^o}CO_2+2H_2O\\C_2H_4+3O_2\xrightarrow[]{t^o}2CO_2+2H_2O$

$n_{CO_2}=n_{CH_4}+2.n_{C_2H_4}=0,05+2.0,05=0,15(mol)$

$m_{CO_2}=0,15.44=6,6(g)$

$n_{O_2}=2.n_{CH_4}+3.n_{C_2H_4}=2.0,05+3.0,05=0,25(mol)$

$V_{kk}=5.V_{O_2}=5.0,25.22,4=28(l)$