1/
a,
Khí thoát ra là metan
=> $V_{C_2H_4}= 22,4-13,44= 8,96l$
=> $n_{C_2H_4}= \frac{8,96}{22,4}= 0,4 mol$
$C_2H_4+ Br_2 \to C_2H_4Br_2$
=> $n_{Br_2}= 0,4 mol$
=> $m_{Br_2}= 160.0,4= 64g$
b,
$\%V_{CH_4}= \frac{13,44.100}{22,4}= 60\%$
$\%V_{C_2H_4}= 40\%$
c,
$n_{CH_4}= \frac{13,44}{22,4}= 0,6 mol$
$CH_4+ 2O_2 \buildrel{{t^o}}\over\to CO_2+ 2H_2O$
=> $n_{O_2}= 1,2 mol$
=> $V_{kk}= 5V_{O_2}= 5.22,4.1,2= 134,4l$
2/ D (xét tỉ lệ mol 1:1)
3/ C