1. Có 4 lọ mất nhãn, mỗi lọ đựng...

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

1.

Từ 4 dung dịch trích ra các mẫu thử, đánh số thứ tự để tiện đối chiếu kết quả thực nghiệm.

\(\eqalign{
& \left\{ \matrix{
C{H_3}COOH \hfill \cr
{C_6}{H_{12}}{O_6} \hfill \cr
{C_2}{H_5}OH \hfill \cr
{C_{12}}{H_{22}}{O_{11}} \hfill \cr} \right.\buildrel {AgN{O_{3/}}N{H_3}} \over
\longrightarrow \left\langle \matrix{
{C_6}{H_{12}}{O_6}: \downarrow Ag \hfill \cr
\left\{ \matrix{
C{H_3}COOH \hfill \cr
{C_2}{H_5}OH \hfill \cr
{C_{12}}{H_{22}}{O_{11}} \hfill \cr} \right.\buildrel { + NaHC{O_3}} \over
\longrightarrow \left\langle \matrix{
C{H_3}COOH: \uparrow C{O_2} \hfill \cr
\left\{ \matrix{
{C_2}{H_5}OH \hfill \cr
{C_{12}}{H_{22}}{O_{11}} \hfill \cr} \right. \hfill \cr} \right. \hfill \cr} \right. \cr
& \left\{ \matrix{
{C_2}{H_5}OH \hfill \cr
{C_{12}}{H_{22}}{O_{11}} \hfill \cr} \right.\buildrel { + Cu{{(OH)}_2}} \over
\longrightarrow \left\langle \matrix{
{C_2}{H_5}OH \hfill \cr
{C_{12}}{H_{22}}{O_{11}}:phuc\,xanh \hfill \cr} \right. \cr
& {C_5}{H_{11}}{O_5}CHO{\rm{ }} + {\rm{ }}A{g_2}O\buildrel {N{H_3}} \over
\longrightarrow N{H_3}{C_5}{H_{11}}{O_5}COOH{\rm{ }} + {\rm{ }}2Ag \downarrow \cr
& C{H_3}COOH{\rm{ }} + {\rm{ }}NaHC{O_3} \to {\rm{ }}C{H_3}COONa{\rm{ }} + {\rm{ }}C{O_2} \uparrow {\rm{ }} + {\rm{ }}{H_2}O \cr} \)

2.

\(\eqalign{
& Ca{C_2} + {\rm{ }}2{H_2}O{\rm{ }} \to {\rm{ }}Ca{\left( {OH} \right)_2} + {\rm{ }}CH \equiv CH \cr
& CH \equiv CH{\rm{ }} + {\rm{ }}{H_2}\buildrel {Pd/to} \over
\longrightarrow C{H_2} = C{H_2} \cr
& C{H_2} = C{H_2} + {\rm{ }}{H_2}O\buildrel {{H_2}S{O_4}} \over
\longrightarrow C{H_3}C{H_2}OH \cr
& C{H_3}C{H_2}OH{\rm{ }} + {\rm{ }}{O_2}\buildrel {xt,{t^o}} \over
\longrightarrow C{H_3}COOH{\rm{ }} + {\rm{ }}{H_2}O \cr
& C{H_3}COOH{\rm{ }} + {\rm{ }}{C_2}{H_5}OH\buildrel {{H_2}S{O_4}} \over
\longrightarrow C{H_3}COO{C_2}{H_5} + {\rm{ }}{H_2}O \cr
& C{H_3}COO{C_2}{H_5} + {\rm{ }}NaOH{\rm{ }} \to {\rm{ }}C{H_3}COONa{\rm{ }} + {\rm{ }}{C_2}{H_5}OH \cr
& CH \equiv CH{\rm{ }} + {\rm{ }}HCl \to C{H_2} = CHCl \cr
& nC{H_2} = CHCl\buildrel {TH} \over
\longrightarrow - {[C{H_2} - CH\left( {Cl} \right)]_n} - {\rm{ }}\left( {PVC:{\rm{ }}polivinylclorua} \right) \cr} \)