1. Iso amylaxetat (thường gọi là...

- Hướng dẫn giải

Phương pháp giải:

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)

\(\eqalign{
& NX\left\{ \matrix{
{n_X} = 0,012 \hfill \cr
{n_{B{r_2}}} = 0,02 \hfill \cr} \right. \to so\,\overline \pi = {{{n_{B{r_2}}}} \over {{n_X}}} = 1,67\buildrel {khong\,co\,khi\,thoat\,ra} \over
\longrightarrow \left\{ \matrix{
1\pi \hfill \cr
> 1\pi \hfill \cr} \right. \cr
& so\,C = {{{n_{C{O_2}}}} \over {{n_X}}} = {{0,032} \over {0,012}} = 2,67 \cr
& {\bf{T}}H1: \cr
& A\left\{ \matrix{
1\pi \hfill \cr
{C_{(A)}} < 2,67 \hfill \cr} \right. \to \left\{ \matrix{
(A){C_2}{H_4}:x \hfill \cr
(B){C_n}{H_{2n + 2 - 2k}}:y \hfill \cr} \right. \to \left\{ \matrix{
\buildrel {BTNT.C} \over
\longrightarrow 2x + ny = 0,032 \hfill \cr
x + y = 0,012 \hfill \cr
\buildrel {BT\,lk\,\pi } \over
\longrightarrow x + ky = 0,02 \hfill \cr} \right. \cr
& = > n = k + 1 = > B:{C_n}{H_4} = > \left[ \matrix{
{C_3}{H_4}:CH \equiv C - C{H_3};C{H_2} = C = C{H_2} \hfill \cr
{C_4}{H_4}:CH \equiv C - CH = C{H_2};C{H_2} = C = C = C{H_2} \hfill \cr} \right. \cr
& TH2: \cr
& A\left\{ \matrix{
1\pi \hfill \cr
{C_{(A)}} > 2,67 \hfill \cr} \right. \to B\left\{ \matrix{
> 1\pi \hfill \cr
{C_{(B)}} < 2,67 \hfill \cr} \right. \to B:{C_2}{H_2} \to \left\{ \matrix{
(A):{C_m}{H_{2m}}:a \hfill \cr
(B){C_2}{H_2}:b \hfill \cr} \right. \cr
& \to \left\{ \matrix{
\buildrel {BTBNT\,C} \over
\longrightarrow am + 2b = 0,032 \hfill \cr
\left. \matrix{
a + b = 0,012 \hfill \cr
\buildrel {BT\,lk\,\pi } \over
\longrightarrow a + 2b = 0,02 \hfill \cr} \right\} \to B\left\{ \matrix{
a = 0,004 = > m = 4 = > A:{C_4}H8 \hfill \cr
b = 0,008 \hfill \cr} \right. \hfill \cr} \right. \cr
& = > \left[ \matrix{
{C_2}{H_4};{C_3}{H_4}/{C_4}{H_4} \hfill \cr
{C_2}{H_2};{C_4}{H_8} \hfill \cr} \right. \cr} \)