$n_A=\dfrac{2,24}{22,4}=0,1(mol)$

$n_{CO_2}=\dfrac{8,96}{22,4}=0,4(mol)$

Đặt CTTQ ankin $A$ là $C_nH_{2n-2}$

Bảo toàn $C$: $n.n_A=n_{CO_2}$

$\to n=\dfrac{n_{CO_2}}{n_A}=4$

Vậy CTPT ankin là $C_4H_6$

CTCT: 

$CH\equiv C-CH_2-CH_3$

$CH_3-C\equiv C-CH_3$

Đáp án:

$\rm C_4H_6$

$\rm But-1-in:\quad CH\equiv C - CH_2 - CH_3$
$\rm But-2-in:\quad CH_3-C\equiv C - CH_3$

Giải thích các bước giải:

\(\begin{array}{l}
n_{ankin} = \dfrac{2,24}{22,4} = 0,1\,\rm mol\\
n_C = n_{\rm CO_2} = \dfrac{8,96}{22,4} = 0,4\, \rm mol\\
n_H = 2n_{\rm H_2O} = 2(n_{\rm CO_2} - n_{ankin}) = 2(0,4 - 0,1) = 0,6\, \rm mol\\
m_{ankin} = m_C + m_H = 0,4.12 + 0,6.1 = 5,4\, \rm gam\\
M_{ankin} = \dfrac{5,4}{0,1} = 54\, \rm g/mol\\
12n + 2n - 2 = 54 \Leftrightarrow n = 4\\
\rm CTPT\,A:\, C_4H_6\\
\rm CTCT\, A:\\
\rm But-1-in:\quad CH\equiv C - CH_2 - CH_3\\
\rm But-2-in:\quad CH_3-C\equiv C - CH_3
\end{array}\)