`#28`

`a.`

PTHH : `C_2H_2+2Br_2->C_2H_2Br_4`

`b.`

`n_{C_2H_2Br_4}=(34,6)/(346)=0,1` `mol`

Theo phương trình : `n_{Br_2}=2n_{C_2H_2Br_4}=0,2` `mol`

`=>m_{Br_2}=0,2.160=32` `g`

`c.`

Theo phương trình : `n_{C_2H_2}=n_{C_2H_2Br_4}=0,1` `mol`

`=>m_{C_2H_2}=0,1.26=2,6` `g`

`=>n_{CH_4}=(7,437)/(24,79)-0,1=0,2` `mol`

`=>m_{CH_4}=0,2.16=3,2` `g`

`=>m_{hh}=3,2+2,6=5,8` `g`

`=>%C_2H_2=(2,6)/(5,8).100%~~44,83%`

`%CH_4=(3,2)/(5,8).100%~~55,17%`

a)

`C_2H_2 + 2Br_2 -> C_2H_2Br_4`

b)`n_{C_2H_2Br_4} = (34,6)/(34,6) = 0,1` mol

`n_{Br_2} = 2n_{C_2H_2Br_4} = 2 . 0,1 = 0,2` mol

`=> m_{Br_2} = 0,2 . 160 = 32g`

c)

`n_{hh khí} = (7,437)/(24,79) = 0,3` mol

`n_{C_2H_2} = n_{C_2H_2Br_4} = 0,1` mol

`=> n_{CH_4} = 0,3 - 0,1 = 0,2` mol

`=>%m_{C_2H_2} = (0,1 . 26)/(0,1 . 26 + 0,2 . 16) . 100% ~~ 44,83%`

`%m_{CH_4} = 100% - 44,83% = 55,17%`