$12)$

$11.2(dm^3)$ hay $0.5(mol)$ phản ứng với $0.625(mol)$ $Br_2.$

`=>`$n_{hh}:n_{Br_2}=\dfrac{4}{5}$

`=>`$n_{Br_2}=\dfrac{5n_{hh}}{4}$

$n_{H_2O}=0.7(mol)$

$m_{hh}=12n_{CO_2}+2n_{H_2O}$

`=>`$n_{CO_2}=0.8(mol)$

Công thức đốt cháy:

$n_{CO_2}-n_{H_2O}=(k-1)n_{hh}=n_{Br_2}-n_{hh}$

`<=>`$0.8-0.7=\dfrac{5n_{hh}}{4}-n_{hh}$

`=>`$n_{hh}=0.4(mol)$

`->`$n_{CH_4}+n_{C_2H_2}+n_{C_3H_6}=0.4$ $(1)$

`=>`$n_{Br_2/11(g) \,\, hh}=0.5(mol)$

Có: $2n_{C_2H_2}+n_{C_3H_6}=0.5$ $(2)$

Lại có: $16n_{CH_4}+26n_{C_2H_2}+42n_{C_3H_6}=11$$(3)$

Từ $(1),(2),(3)$ suy ra:

$n_{CH_4}=0.1(mol);n_{C_2H_2}=0.2(mol);n_{C_3H_6}=0.1(mol)$

$\%V_{CH_44}=\%V_{C_3H_6}=25\%||\%V_{C_2H_2}=50\%$

TN1:

`Hh:C_2H_2:x(mol);C_3H_6:y(mol);CH_4:z(mol)`

`->26x+42y+16z=11(1)`

BTNT H: `n_{H_2O}=x+3y+2z={12,6}/{18}=0,7(2)`

TN2:

`Hh:C_2H_2:kx(mol);C_3H_6:ky(mol);CH_4:kz(mol)`

`->k(x+y+z)={11,2}/{22,4}=0,5(3)`

`C_2H_2+2Br_2->C_2H_2Br_4`

`C_3H_6+Br_2->C_3H_6Br_2`

`->n_{Br_2}=2kx+ky={100}/{160}=0,625(4)`

`(3)(4)->{x+y+z}/{2x+y}=0,8`

`->0,6x-0,2y-z=0(5)`

`(1)(2)(5)->x=0,2;y=0,1;z=0,1`

`->\%V_{C_2H_2}=50\%;\%V_{C_3H_6}=\%V_{CH_4}=25\%`