1. Hợp chất hữu cơ A chỉ chứa 2 n...

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

1.

\(\eqalign{
& \left\{ \matrix{
A\buildrel { + {O_2}} \over
\longrightarrow {H_2}O \to A\,co\,H \hfill \cr
A:hchc\, = > \,A\,co\,C \hfill \cr} \right. \to A:{C_n}{H_{2n + 2 - 2k}}(n,k \in N*) \cr
& {C_n}{H_{2n + 2 - 2k}} + {{3n + 1 - k} \over 2}{O_2} \to nC{O_2} + (n + a - k){H_2}O \cr
& {m_A} = {m_{{H_2}O}} \to \left\{ \matrix{
14n + 2 - 2k = 18(n + 1 - k) \hfill \cr
= > 4n - 16k + 16 = 0 \hfill \cr} \right. = > n + 4 = 4k = > (k;n) = \left[ \matrix{
(2,4) \to {C_4}{H_6} \hfill \cr
(3,8) \to {C_8}{H_{12}} \hfill \cr
(4,12) \to {C_{12}}{H_{18}} \hfill \cr} \right. \cr} \)

2.

\(\eqalign{
& \left\{ \matrix{
{C_6}{H_{12}}{O_6} + \mathop {6{O_2}}\limits_1 \to \mathop {6C{O_2}}\limits_1 + 6{H_2}O \hfill \cr
{C_{12}}{H_{22}}{O_{11}} + \mathop {12{O_2}}\limits_1 \to \mathop {12C{O_2}}\limits_1 + 11{H_2}O \hfill \cr} \right.\buildrel {NX} \over
\longrightarrow \left\{ \matrix{
{n_{{O_2}}} = {n_{C{O_2}}} \hfill \cr
= > {n_{C{O_2}}} = 0,6 \hfill \cr} \right. \cr
& \buildrel { + Ba{{(OH)}_2}\,du} \over
\longrightarrow \left\{ \matrix{
n \downarrow = {n_{C{O_2}}} = 0,6 \hfill \cr
= > m \downarrow = 118,2g \hfill \cr} \right. \cr} \)