Em tham khảo nha:
\(\begin{array}{l}
a)\\
CTTQ\,X:{C_n}{H_{2n - 2}}\\
{M_X} = 20 \times 2 = 40g/mol\\
\Rightarrow 14n - 2 = 40 \Leftrightarrow n = 2\\
CTPT:{C_3}{H_4}(propin)\\
b)\\
\% {m_H} = 100 - 88,24 = 11,76\% \\
x:y = \dfrac{{88,24}}{{12}}:\dfrac{{11,76}}{1} = 7,35:11,76 = 5:8\\
\Leftrightarrow CTPT:{C_5}{H_8}\\
c)\\
CTTQ\,X:{C_n}{H_{2n - 2}}\\
{C_n}{H_{2n - 2}} + 2B{r_2} \to {C_n}{H_{2n - 2}}B{r_4}\\
{n_{B{r_2}}} = \dfrac{{64}}{{160}} = 0,4\,mol\\
{n_X} = \dfrac{{0,4}}{2} = 0,2\,mol \Rightarrow {M_X} = \dfrac{{10,8}}{{0,2}} = 54g/mol\\
\Rightarrow 14n - 2 = 54 \Leftrightarrow n = 4\\
CTPT:{C_4}{H_6}\\
d)\\
CH \equiv C - C{H_3} + AgN{O_3} + N{H_3} \to CAg \equiv C - C{H_3} + N{H_4}N{O_3}\\
{n_{{C_3}{H_4}}} = \dfrac{8}{{40}} = 0,2\,mol\\
{n_{{C_3}{H_3}Ag}} = {n_{{C_3}{H_4}}} = 0,2\,mol\\
{m_{{C_3}{H_3}Ag}} = 0,2 \times 147 = 29,4g\\
g)\\
{n_{C{O_2}}} = \dfrac{{13,2}}{{44}} = 0,3\,mol\\
{n_{{H_2}O}} = \dfrac{{3,6}}{{18}} = 0,2\,mol\\
{n_X} = {n_{C{O_2}}} - {n_{{H_2}O}} = 0,3 - 0,2 = 0,1\,mol\\
n = \dfrac{{{n_{C{O_2}}}}}{{{n_X}}} = \dfrac{{0,3}}{{0,1}} = 3 \Rightarrow CTPT:{C_3}{H_4}
\end{array}\)
