1) $M_M = M_{CH_4}.3,625 = 16.3,625 = 58(g/mol)$
Suy ra : M là $C_4H_{10}$
$C_4H_{10} + \dfrac{5}{2}O_2 \xrightarrow{t^o,xt} 2CH_3COOH + H_2O$
$C_4H_{10} \xrightarrow{xt,t^o} CH_4 + C_3H_6$
$CH_3COOH + 2NaOH \xrightarrow{CaO,t^o} CH_4 + Na_2CO_3 + H_2O$
$M : C_4H_{10} ; N : CH_3COOH ; P : CH_4$
2) Coi A là $C_nH_4$
Ta có : $M_A = 12n + 4 = 17.2 = 34 \Rightarrow n = 2,5$
$n_C = n_{CO_2} = n_{CaCO_3} = 0,3(mol)$
$\Rightarrow n_A = \dfrac{0,3}{2,5} = 0,12(mol)$
$\Rightarrow m_A = 0,12.34 = 4,08(gam)$
Độ bất bão hoà của A : $k = \dfrac{C.2 + 2 -H}{2} = \dfrac{2,5.2 + 2 - 4}{2} = 1,5$
Ta có: $n_{Br_2} = k.n_A \Rightarrow 0,27 = 1,5.a \Rightarrow a = 0,18(mol)$
