`X:`$\begin{cases} C_3H_6: x(mol)\\C_4H_10: y(mol)\\C_2H_2: z(mol)\\H_2: t(mol)\end{cases}$
Bảo toàn nguyên tố `H`:
`n_{H_2O}= 3x + 5y + z + t= 0,675(mol)`
Bảo toàn nguyên tố `C`:
`n_{CO_2}= 3x + 4y + 2z(mol)`
`n_{Br_2}= {150}/{1000}. 1= 0,15(mol)`
`n_{Br_2}= {64}/{160}= 0,4(mol)`
`n_X = {11,2}/{22,4}= 0,5(mol)`
`11,2l` `X` có: $\begin{cases} C_3H_6: kx(mol)\\C_4H_10: ky(mol)\\C_2H_2: kz(mol)\\H_2: kt(mol)\end{cases}$
Bảo toàn liên kết `pi`:
`n_{Br_2} = n_{C_3H_6} + 2n_{C_2H_2}`
`= k(x + 2z)(mol)`
`{n_X}/{n_{Br_2}}= {k(x + y + z + d)}/{k(x+2z)}= {0,5}/{0,4}= 5/4`
`<=> x + 6z - 4y - 4t = 0(a)`
Bảo toàn liên kết `pi`:
`n_{C_3H_6} + 2n_{C_2H_2}= n_{Br_2} + n_{H_2}`
`<=> x + 2z = 0,15 + t`
`<=> x + 2z - t = 0,15(mol)(b)`
Lấy `4b - a`:
`4(x+ 2z - t) - (x+ 6z - 4y - 4t) = 0,15. 4-0`
`<=> 3x + 2z+ 4y = 0,6(mol)= n_{CO_2}`
`V= 0,6. 22,4 = 13,44(l)`