1.
$Al_4C_3+12H_2O\to 4Al(OH)_3+3CH_4$
$2CH_4\xrightarrow{{\text{ lln}, 1500^oC}} C_2H_2+3H_2$
$2C_2H_2\xrightarrow{{CuCl, NH_4Cl, t^o}} CH\equiv C-CH=CH_2$
$CH\equiv C-CH=CH_2+H_2\xrightarrow{{Pd/PbCO_3, t^o}} CH_2=CH-CH=CH_2$
$nCH_2=CHCH=CH_2\xrightarrow{{t^o, p, xt}} (-CH_2-CH=CH-CH_2-)_n$ (cao su Buna)
2.
$n_{CO_2}=n_{CaCO_3\downarrow}=\dfrac{30}{100}=0,3(mol)$
$m_{\text{tăng}}=16,8g=m_{CO_2}+m_{H_2O}$
$\to n_{H_2O}=\dfrac{16,8-0,3.44}{18}=0,2(mol)$
Gọi CTTQ ankin là $C_nH_{2n-2}$
$n_Y=n_{CO_2}-n_{H_2O}=0,3-0,2=0,1(mol)$
$\to n=\dfrac{n_{CO_2}}{n_Y}=3$
Vậy CTPT ankin là $C_3H_4$