Đáp án+Giải thích các bước giải:

`a.`

`CH_{2}=CH_{2}+Br_{2}→CH_{2}Br-CH_{2}Br`

`CH≡CH+2Br_{2}→CHBr_{2}-CHBr_{2}`

`b.`

Đặt `n_{C_{2}H_{4}}=x(mol)`

      `n_{C_{2}H_{2}}=y(mol)`

`n_{hh}=x+y=(0,56)/(22,4)=0,025(1)`

`n_{Br_{2}}=x+2y=(5,6)/(160)=0,035(2)`

`(1)(2)→x=0,015;y=0,01`

`→%V_{C_{2}H_{4}}=(0,015.22,4)/(0,56).100%=60(%)`

`→%V_{C_{2}H_{2}}=100-60=40(%)`

`c.`

`PTHH(1):n_{C_{2}H_{4}Br_{2}}=n_{C_{2}H_{4}}=0,015(mol)`

`PTHH(2):n_{C_{2}H_{2}Br_{4}}=n_{C_{2}H_{2}}=0,01(mol)`

$m_{muối}=m_{C_{2}H_{4}Br_{2}}+m_{C_{2}H_{2}Br_{4}}=0,015.188+0,01.346=6,28(g)$

$m_{khí }=m_{muối}-m_{Br_{2}}=6,28-5,6=0,68(g)$

`→M_{A}=(0,68)/(0,025)=27,2` $(g/mol)$