Cho ankin có công thức là : `C_nH_{2n-2}`

`a)` Ta có: :

`M_{C_nH_{2n-2}}=M_{(kk)}.0,89655`

`M_{C_nH_{2n-2}}=29.0,89655`

`M_{C_nH_{2n-2}}=26g`/`mol`

`\to 12n+2n-2=26`

`\to 14n=28`

`\to n=2`

Thế `n` vào vào công thức.

Vậy `\text{CTPT}` của ankin trên là: `C_2H_2` `(\text{ axetilen})`.

`b)` `n_{CH_4}=\frac{4,48}{22,4}=0,2(mol)`

Ta có: `∑_{\text{hh}}=\frac{m_{CH_4}+m_{C_2H_2}}{n_{CH_4}+n_{C_2H_2}}`

`=> ∑_{\text{hh}}=\frac{16.0,2+26.0,1}{0,2+0,1}`

`=> ∑_{\text{hh}}=\frac{58}{3}`

$d_{\text{X}/\text{kk}}=\dfrac{M_X}{M_{\text{kk}}}=\dfrac{\dfrac{58}{3}}{29}$

$d_{\text{X}/\text{kk}}=\dfrac{2}{3}$

`c)`  Cho `n_{C_2H_2}:n_{CH_4}:n_{C_2H_4}=1:2:1=x`

Ta có: `n_{C_2H_2}=x`

`n_{CH_4}=2x`

`n_{C_2H_4}=x`

Do: `m_{C_2H_2}+m_{CH_4}+m_{C_2H_4}=12,9g`

`=> 26x+2.16x+28x=12,9`

`=> x=0,15(mol)`

`=>` `n_{\text{hh}}=0,15+0,15.2+0,15=0,6(mol)`

Ta có: ` ∑_{\text{hh}}=\frac{m_{\text{hh}}}{n_{\text{hh}}}=\frac{12,9}{0,6}=21,5g`/`mol`

`=>` $d_{\text{Y}/O_2}=\dfrac{21,5}{32}=0,671875$

Khối lượng mỗi chất trong `Y` là:

`m_{C_2H_2}=26x=26.0,15=3,9g`

`m_{C_2H_4}=28x=28.0,15=4,2g`

`m_{CH_4}=32x=32.0,15=4,8g`

 

a,

$M=0,89655.29=26$

Đặt CTTQ ankin là $C_nH_{2n-2}$

$\to 12n+2n-2=26$

$\to n=2$

Vậy CTPT $X$ là $C_2H_2$

CTCT: $HC\equiv CH$ (axetilen)

b,

$n_{CH_4}=\dfrac{4,48}{22,4}=0,2(mol)$

$\to\overline{M}_X=\dfrac{0,1.26+0,2.16}{0,1+0,2}=\dfrac{58}{3}$

$\to d_{X/kk}=\dfrac{58}{3}:29=\dfrac{2}{3}$

c,

Xét $0,6$ mol hỗn hợp $Y$

Đặt $n_{C_2H_2}=a, n_{CH_4}=2a, n_{C_2H_4}=a(mol)$

$\to a+2a+a=0,6$

$\to a=0,15$

$\to n_{C_2H_2}=n_{C_2H_4}=0,15(mol); n_{CH_4}=0,3(mol)$

$\overline{M}_Y=\dfrac{0,15.26+0,3.16+0,15.28}{0,6}=21,5$

$\to d_{Y/O_2}=0,671875$

Vì $0,15.26+0,3.16+0,15.28=12,9$

Nên trong $12,9g$ hh $Y$ có:

$m_{C_2H_2}=0,15.26=3,9g$

$m_{CH_4}=0,3.16=4,8g$

$m_{C_2H_4}=0,15.28=4,2g$