Câu 6 trang 39 SGK Hóa 12 Nâng cao

Thủy phân hoàn toàn \(34,2\) g saccarozơ, sau đó tiến hành phản ứng tráng bạc với dung dịch thu được. Tính khối lượng \(Ag\) kết tủa.

Giải

\(n_{{C_{12}}{H_{22}}{O_{11}}} = {{34,2} \over {342}} = 0,1(mol) \)

\({C_{12}}{H_{22}}{O_{11}} + {H_2}O\buildrel {{H_2}S{O_{4,}}t^\circ } \over
\longrightarrow {C_6}{H_{12}}{O_6} + {C_6}{H_{12}}{O_6}\)

Saccarozơ                                 glucozơ         fructozơ

\(\eqalign{
& 0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\buildrel {} \over
\longrightarrow \,\,\,\,\,\,0,1\,\,\,\,\,\,\,\,\,\,\,\,\buildrel {} \over
\longrightarrow 0,1 \cr
& \cr} \)

Trong môi trường kiềm fructozơ tồn tại cân bằng sau:

\(\eqalign{
& C{H_2}OH - {(CHOH)_3} - C - C{H_2}OH \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} C{H_2}OH - {(CHOH)_4} - CHO\,\,\,\,(1) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\| {} \right. \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O \cr} \)

Glucozơ tham gia phản ứng tráng gương làm cho cân bằng (1) chuyển dịch theo chiều thuận.

\(\eqalign{
& C{H_2}OH{(CHOH)_4}CHO \to 2Ag \downarrow \cr
& 0,1 + 0,1 = 0,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to 2.0,2 = 0.4 \cr
& \Rightarrow {m_{Ag}} = 0,4.108 = 43,2(g) \cr} \)