Đáp án đúng:
Giải chi tiết:1.
\(\matrix{ {C{H_3}COOH{\rm{ }} + {\rm{ }}{C_5}{H_{11}}OH{\rm{ }} \to {\rm{ }}C{H_3}COO{C_5}{H_{11}} + {\rm{ }}{H_2}O} \hfill \cr \matrix{ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,15 \hfill \cr \buildrel {H = 80\% } \over \longrightarrow {n_{C{H_3}COOH}} = {n_{{C_5}{H_{11}}OH}} = {{0,15} \over {80\% }} = 0,1875 \hfill \cr = > \left\{ \matrix{ {m_{C{H_3}COOH}} = 11,25g \hfill \cr {m_{{C_5}{H_{11}}OH}} = 16,5g \hfill \cr} \right. \hfill \cr} \hfill \cr} \)
2.
X là hỗn hợp khí nên số C ≤ 4 (-C5H12)
\(\eqalign{ & NX\left\{ \matrix{ {n_X} = 0,012 \hfill \cr {n_{B{r_2}}} = 0,02 \hfill \cr} \right. \to so\,\overline \pi = {{{n_{B{r_2}}}} \over {{n_X}}} = 1,67\buildrel {khong\,co\,khi\,thoat\,ra} \over \longrightarrow \left\{ \matrix{ 1\pi \hfill \cr > 1\pi \hfill \cr} \right. \cr & so\,C = {{{n_{C{O_2}}}} \over {{n_X}}} = {{0,032} \over {0,012}} = 2,67 \cr & {\bf{T}}H1: \cr & A\left\{ \matrix{ 1\pi \hfill \cr {C_{(A)}} < 2,67 \hfill \cr} \right. \to \left\{ \matrix{ (A){C_2}{H_4}:x \hfill \cr (B){C_n}{H_{2n + 2 - 2k}}:y \hfill \cr} \right. \to \left\{ \matrix{ \buildrel {BTNT.C} \over \longrightarrow 2x + ny = 0,032 \hfill \cr x + y = 0,012 \hfill \cr \buildrel {BT\,lk\,\pi } \over \longrightarrow x + ky = 0,02 \hfill \cr} \right. \cr & = > n = k + 1 = > B:{C_n}{H_4} = > \left[ \matrix{ {C_3}{H_4}:CH \equiv C - C{H_3};C{H_2} = C = C{H_2} \hfill \cr {C_4}{H_4}:CH \equiv C - CH = C{H_2};C{H_2} = C = C = C{H_2} \hfill \cr} \right. \cr & TH2: \cr & A\left\{ \matrix{ 1\pi \hfill \cr {C_{(A)}} > 2,67 \hfill \cr} \right. \to B\left\{ \matrix{ > 1\pi \hfill \cr {C_{(B)}} < 2,67 \hfill \cr} \right. \to B:{C_2}{H_2} \to \left\{ \matrix{ (A):{C_m}{H_{2m}}:a \hfill \cr (B){C_2}{H_2}:b \hfill \cr} \right. \cr & \to \left\{ \matrix{ \buildrel {BTBNT\,C} \over \longrightarrow am + 2b = 0,032 \hfill \cr \left. \matrix{ a + b = 0,012 \hfill \cr \buildrel {BT\,lk\,\pi } \over \longrightarrow a + 2b = 0,02 \hfill \cr} \right\} \to B\left\{ \matrix{ a = 0,004 = > m = 4 = > A:{C_4}H8 \hfill \cr b = 0,008 \hfill \cr} \right. \hfill \cr} \right. \cr & = > \left[ \matrix{ {C_2}{H_4};{C_3}{H_4}/{C_4}{H_4} \hfill \cr {C_2}{H_2};{C_4}{H_8} \hfill \cr} \right. \cr} \)
