Đáp án + Giải thích các bước giải:

$a\big)$

Đặt CTPT `X:C_nH_{2n-2}`

`n_{CO_2}={16,8}/{22,4}=0,75(mol)`

Bảo toàn C: `n_X.n=n_{CO_2}`

`->{10,2n}/{14n-2}=0,75`

`->n=5`

Vậy CTPT ankin là `C_5H_8`

$b\big)$

$CH\equiv C-CH_2-CH_2-CH_3:\rm pent-1-in$

$CH_3-C\equiv C-CH_2-CH_3:\rm pent-2-in$

$CH\equiv C-CH(CH_3)_2:\rm 3-metylbut-1-in$

Em tham khảo nha :

\(\begin{array}{l}
a)\\
CTPT:{C_n}{H_{2n - 2}}\\
{n_{C{O_2}}} = \dfrac{{16,8}}{{22,4}} = 0,75mol\\
{n_{{C_n}{H_{2n - 2}}}} = \dfrac{{{n_{C{O_2}}}}}{n} = \dfrac{{0,75\,}}{n}mol\\
{M_X} = 10,2:\dfrac{{0,75}}{n} = 13,6n\\
 \Rightarrow 14n - 2 = 13,6n\\
 \Rightarrow n = 5\\
CTPT:{C_5}{H_8}\\
b)\\
CH \equiv C - C{H_2} - C{H_2} - C{H_3}:Pent - 1 - in\\
C{H_3} - C \equiv C - C{H_2} - C{H_3}:Pent - 2 - in\\
CH \equiv C - CH(C{H_3}) - C{H_3}:3 - metylbut - 1 - in
\end{array}\)