Đáp án:

\(\begin{array}{l}
2\,D\\
3\,A\\
4\,D
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
2)\\
{n_{C{O_2}}} = \dfrac{{8,8}}{{44}} = 0,2\,mol\\
{n_{{H_2}O}} = \dfrac{{4,5}}{{18}} = 0,25\,mol\\
{n_X} = 0,25 - 0,2 = 0,05\,mol\\
n = \dfrac{{0,2}}{{0,05}} = 4 \Rightarrow CTPT:{C_4}{H_{10}}\\
3)\\
C{H_3}COONa + NaOH \to N{a_2}C{O_3} + C{H_4}\\
{n_{C{H_3}COONa}} = \dfrac{{16,4}}{{82}} = 0,2\,mol\\
{n_{C{H_4}}} = {n_{C{H_3}COONa}} = 0,2\,mol\\
H = 80\%  \Rightarrow {V_{C{H_4}}} = 0,2 \times 22,4 \times 80\%  = 3,584l\\
4)\\
A{l_4}{C_3} + 12HCl \to 4AlC{l_3} + 3C{H_4}\\
{n_{C{H_4}}} = \dfrac{{8,4}}{{22,4}} = 0,375\,mol\\
{n_{A{l_4}C{l_3}}} = \dfrac{{0,375}}{3} = 0,125\,mol\\
H = 90\%  \Rightarrow {m_{A{l_4}{C_3}}} = 0,125 \times 144 \times \dfrac{{100}}{{90}} = 20g
\end{array}\)