Bài 1.
Các phản ứng xảy ra:
\(C{H_3}C{H_3} + C{l_2}\xrightarrow{{as}}C{H_3}C{H_2}Cl + HCl\)
\(CH \equiv CH + 2B{r_2}\xrightarrow{{}}CHB{{\text{r}}_2}CHB{{\text{r}}_2}\)
\(CH \equiv CH + 2{H_2}\xrightarrow{{Ni,{t^o}}}C{H_3}C{H_3}\)
Bài 2.
Sơ đồ phản ứng:
\(A + {O_2}\xrightarrow{{{t^o}}}C{O_2} + {H_2}O\)
Ta có:
\({n_{C{O_2}}} = \dfrac{{13,2}}{{44}} = 0,3{\text{ mol = }}{{\text{n}}_C}\)
\({n_{{H_2}O}} = \dfrac{{7,2}}{{18}} = 0,4{\text{ mol > }}{{\text{n}}_{C{O_2}}}\)
Vậy \(A\) là ankan
\( \to {n_A} = {n_{{H_2}O}} - {n_{C{O_2}}} = 0,4 - 0,3 = 0,1{\text{ mol}}\)
\( \to {C_A} = \dfrac{{{n_C}}}{{{n_A}}} = \dfrac{{0,3}}{{0,1}} = 3\)
Suy ra \(A\) là \(C_3H_8\)
\( \to m = {m_A} = 0,1.(12.3 + 8) = 4,4{\text{ gam}}\)