Đáp án:
$\rm H = 77,5\%$
$\rm m_{PE}= 21,7\, gam$
Giải thích các bước giải:
$\rm nCH_2=CH_2 \xrightarrow{\ t^\circ,\ xt,\ p\ } (\kern-6pt- CH_2-CH_2-\kern-6pt)_n$
$\rm CH_2=CH_2 + Br_2 \longrightarrow C_2H_4Br_2$
$\rm n_{C_2H_4\, dư}= n_{Br_2}=\dfrac{36}{160}=0,225\, mol$
$\rm n_{C_2H_4\,pứ}= 1 - 0,225 = 0,775\,mol$
$\rm H =\dfrac{0,775}{1}\times 100\% = 77,5\%$
$\rm m_{PE}= m_{C_2H_4\,pứ}= 0,775\times 28 = 21,7\, gam$