\(1.\)

\(n_C^{} = n_{CO_2^{}}^{} = \frac{{2,688}}{{22,4}} = 0,12\;{\text{mol}}\)

\({\text{m}}_A^{} = m_C^{} + m_H^{} = 1,72\)

\( \to m_H^{} = 1,72 - 0,12.12 = 0,28\;{\text{mol}}\)

\({\text{n}}_H^{} = \frac{{0,28}}{1} = 0,28\;{\text{mol}}\)

Vì A là một hidrocacbon nên A là \({\text{C}}_x^{}H_y^{}\)

\(C:H = 0,12:0,28 = 3:7\)

\( \to CTDG:(C_3^{}H_7^{})_n^{}\)

Vì số hidro trong hidrocacbon luôn chẵn

\( \to C_6^{}H_{14}^{}\)

\(2.n_{Ag}^{} = \frac{{8,64}}{{108}} = 0,08\;{\text{mol}}\)

\({\text{n}}_{hh}^{} = \frac{1}{2}n_{Ag}^{} = 0,04\;{\text{mol}}\)

\(M_{TrungBinh}^{} = \frac{{2,46}}{{0,04}} = 61,5g/mol\)

\( \to C_3^{}H_6^{}O\& C_4^{}H_8^{}O\)

\(3.\)

\((CH_2^{} = CH - C(CH_3^{}) = CH_2^{})_n^{}\xrightarrow{{xt,p,{t^o}}}n( - CH_2^{} - CH = C(CH_3^{}) - CH_2^{} - )\)

\(CH_2^{} = CH - CH_3^{} + HCl\xrightarrow{{H{g^{2 + }}}}CH_3^{} - CHCl - CH_3^{}\)

\(4.\)

\(3HCHO + 4AgNO_3^{} + 6NH_3^{} + 2H_2^{}O\xrightarrow{{{t^o}}}(NH_4^{})_2^{}CO_3^{} + 4Ag + 4NH_4^{}NO_3^{}\)

\(C_6^{}H_5^{}OH + Br_2^{} \to C_6^{}H_2^{}Br_3^{}OH + HBr\)