1. Ba chất hữu cơ mạch hở A, B, C...

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

1.

B tác dụng được với Na và NaOH => B là axit C₂H₅COOH – C₃H₆O₂

A: C₃H₇OH

C: C₂H₅COOC₃H₇

C₂H₅COOH + C₃H₇OH → C₂H₅COOC₃H₇ + H₂O

C₂H₅COOH + Na → C₂H₅COONa + ½ H₂↑

C₃H₇OH + Na → C₃H₇ONa + ½ H₂↑

2.

\(\eqalign{
& {({C_6}{H_{10}}{O_5})_n}\buildrel { + {H_2}O} \over
\longrightarrow {C_6}{H_{12}}{O_6}\buildrel {men} \over
\longrightarrow 2{C_2}{H_5}OH \cr
& \left\{ \matrix{
{C_2}{H_5}OH:50.36\% = 18(l) \to m = D.V = 14,4kg \hfill \cr
{H_2}O:50 - 18 = 32(l) \to m = D.V = 32kg \hfill \cr} \right. \to mglu = {{14,4\,.\,162} \over {46\,.\,2\,.\,80\% \,.\,80\% }} = 39,62kg \cr} \)

\(\eqalign{
& A\left\{ \matrix{
Fe:x \hfill \cr
Cu:y \hfill \cr} \right.\buildrel { + HN{O_3}:0,7mol} \over
\longrightarrow \left\{ \matrix{
\uparrow B:V \hfill \cr
{\rm{dd}}\,X\buildrel { + KOH:0,6} \over
\longrightarrow \left\{ \matrix{
\downarrow Y\buildrel {{t^o}} \over
\longrightarrow ran\,\left\{ \matrix{
F{e_2}{O_3} \hfill \cr
CuO \hfill \cr} \right. \hfill \cr
{\rm{dd}}\,Z\buildrel {{t^o}} \over
\longrightarrow ran\,T\buildrel {{t^o}} \over
\longrightarrow ran\,G:46,65g \hfill \cr} \right. \hfill \cr} \right. \cr
& \left\{ \matrix{
Fe:x \hfill \cr
Cu:y \hfill \cr} \right. \to \left\{ \matrix{
F{e_2}{O_3}:0,5x \hfill \cr
CuO:y \hfill \cr} \right. \to \left\{ \matrix{
56x + 64y = 11,6 \hfill \cr
160\,.\,0,5x + 80y = 16 \hfill \cr} \right. \to \left\{ \matrix{
x = 0,15 \hfill \cr
y = 0,05 \hfill \cr} \right. \cr
& G\left\{ \matrix{
KOH\,du:a \hfill \cr
KN{O_2}:b \hfill \cr} \right. \to \left\{ \matrix{
56a + 85b = 46,65 \hfill \cr
\buildrel {BTNT\,K} \over
\longrightarrow a + b = 0,6 \hfill \cr} \right. \to \left\{ \matrix{
a = 0,15 \hfill \cr
b = 0,45 \hfill \cr} \right. \to {\rm{dd}}\,Z\left\{ \matrix{
KOH\,du:0,15 \hfill \cr
KN{O_3}:0,45 \hfill \cr} \right. \cr
& X\left\{ \matrix{
Fe{(N{O_3})_2}:c \hfill \cr
Fe(N{O_3})3:b \hfill \cr
Cu{(N{O_3})_2}:0,05 \hfill \cr} \right. \to \left\{ \matrix{
\buildrel {BTNT\,Fe} \over
\longrightarrow c + d = 0,15 \hfill \cr
\buildrel {BT\,N{O_3}^ - } \over
\longrightarrow 2c + 3d + 2.0,05 = 0,45 \hfill \cr} \right. \to \left\{ \matrix{
c = 0,1 \hfill \cr
d = 0,05 \hfill \cr} \right. \cr
& \left\{ \matrix{
\buildrel {BTNT\,\,N} \over
\longrightarrow {n_{N( \uparrow )}} = {n_{HN{O_3}}} - {n_{N{O_3}(X)}} = 0,25 \hfill \cr
\buildrel {BTNT\,H} \over
\longrightarrow {n_{HCl}} = 2{n_{{H_2}O}} = 0,7\buildrel {BTNT\,O} \over
\longrightarrow \left\{ \matrix{
3{n_{HN{O_3}}} = {n_{O( \uparrow )}} + 3{n_{N{O_3}(X)}} + {n_{{H_2}O}} \hfill \cr
= > {n_{O( \uparrow )}} = 0,4 \hfill \cr} \right. \hfill \cr} \right. \cr} \)

=> nNO = 0,1 và nNO2 = 0,15

BTKL

m_X + m_{dd HNO3} = m_{dd X} + m_H2O + m↑

=> m_{dd X} = 11,6 + 87,5 – 30 . 0,1 – 46 . 0,15 = 89,2g

=> C%Fe(NO₃)₃ = 13,565%