Cho `2` ancol có CT là `C_\overline{n}H_{2\overline{n}+1}OH` `(n\ge 0)`
`n_{H_2}=\frac{1,68}{22,4}=0,075(mol)`
`2C_{\overline{n}}H_{2\overline{n}+1}OH+2Na\to2C_\overline{n}H_{2\overline{n}+1}ONa+H_2 `
`n_{C_\overline{n}H_{2\overline{n}+1}OH}=2n_{H_2}`
`=>n_{C_\overline{n}H_{2\overline{n}+1}OH}=0,15(mol)`
`=>\overline{M}_{ancol}=\frac{6}{0,15}=40g`/`mol`
$ 14\overline{n}+18=40$
$\to \overline{n}\approx 1,5$
Do đồng đẳng liên tiếp nên `2` ancol có CTPT là: $\begin{cases} CH_3OH\\ C2H_5OH\\\end{cases}$
Cho $\begin{cases} CH_3OH:x(mol)\\ C_2H_5OH:y(mol)\\\end{cases}$
`=>`$\begin{cases}x+y=0,15\\32x+46y=6g\\\end{cases}$`<=>`$\begin{cases}x=\dfrac{9}{140}(mol)\\y=\dfrac{3}{35}(mol)\\\end{cases}$
`%m_{CH_3OH}=\frac{9.32.100%}{140.6}\approx 34,28%`
`%m_{C_2H_5OH}=\frac{3.46.100%}{35.6}\approx 65,7%`
