$n_{CO_2}=0.7(mol)$

Quy hỗn hợp:

$\begin{cases} O:a\\CH_2:0.5\\CH_4:0.2 \end{cases}$

Có : $\%m_O=7.27\%$

`->`$\dfrac{16a}{16a+14 \times 0.5+0.2 \times 16}=0.0727$

`=>`$a=0.05$

Bảo toàn $e:$

$-2a+6 \times 0.5+0.2 \times 8=4n_{O_2}$

`=>`$n_{O_2}=\dfrac{-2 \times 0.05+6 \times 0.5+0.2 \times 8}{4}=1.125(mol)$

`=>`$V_{O_2}=1.125 \times 22.4=25.2(L)$

Sửa đề: `CO -> CO_2`

Hỗn hợp Z:

`CH_3OH`

`C_3H_8`

`C_4H_{10}`

`C_5H_{12}`

Có:

`C_3H_8 -> CH_4 + 2CH_2`

`C_4H_{10} -> CH_4 + 3CH_2`

`C_5H_{12} -> CH_4 + 4CH_2`

Nên quy đổi Z thành:

`CH_3OH: x(mol)`

`CH_4: y(mol)`

`CH_2: z(mol)`

`=> x + y = 0,2(1)`

`\%m_O = (16x)/(32x + 16y + 14z) .100\% = 7,27\%(2)`

`n_{CO_2} = (30,8)/(44) = 0,7(mol)`

Bảo toàn C: `n_{CO_2} = n_{CH_3OH} + n_{CH_4} + n_{CH_2}`

`=> x + y + z = 0,7(3)`

`(1),(2),(3) => x = 0,05; y = 0,15; z = 0,5`

Bảo toàn H: `n_{H_2O} = 2n_{CH_3OH} + 2n_{CH_4} + n_{CH_2} = 0,9(mol)`

Bảo toàn O: `n_{O_2} = n_{CO_2} + 1/2 n_{H_2O} -1/2 n_{CH_3OH} = 1,125(mol)`

`=> V = 1,125.24,79 = 27,88875(l)`