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Đáp án:

`H%=40%`

Giải thích các bước giải:

`C_4H_{10} \to CH_4 + C_3H_6`        (1)

`C_4H_{10} \to C_2H_6 + C_2H_4`      (2)

Đặt `n_{C_4H_{10}(1)=a mol`

       `n_{C_4H_{10}(2)=b mol`

       `n_{C_4H_{10}.dư}= c mol`

Theo pt:

`m_{ankan}=16a+30b+58c=4,54g`      (3)

`n_{anken}=a+b=0,04 mol`                  (4)

`CH_4 + 2O_2 \to CO_2 + H_2O`

`2C_2H_6 + 7O_2 \to 4CO_2 + 6H_2O`

`2C_4H_{10} + 13O_2 \to 8CO_2 + 10H_2O`

`=> n_{O_2}=2a+3,5b+6,5c=0,515 mol`      (5)

Giải hpt (3)(4)(5) `=> a=0,01`    `b=0,03`    `c=0,06`

`=> H%={0,01+0,03}/{0,01+0,03+0,06}.100=40%`

$n_{\text{anken}}=n_{C_4H_{10}\text{pứ}}$

Gọi chung các ankan trong Y là $C_nH_{2n+2}$

$n_Y=\dfrac{4,54}{14n+2}(mol)$

$C_nH_{2n+2}+\dfrac{3n+1}{2}O_2\to nCO_2+(n+1)H_2O$

$\Rightarrow \dfrac{4,54(3n+1)}{2.(14n+2)}=0,515$

$\Leftrightarrow 13,62n+4,54=0,515.2(14n+2)$

$\Leftrightarrow n=3,1$

$\Rightarrow n_Y=\dfrac{4,54}{14n+2}=0,1(mol)$

$\Rightarrow n_{C_4H_{10}\text{bđ}}=0,1(mol)$

$\to H=\dfrac{0,04.100}{0,1}=40\%$