Hỗn hợp rắn X gồm M, MO và MCl2 (M là k...

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

\(X\left\{ \matrix{
M \hfill \cr
MO \hfill \cr
MC{l_2} \hfill \cr} \right. \to \left[ \matrix{
\buildrel {HCl\,vua\,du} \over
\longrightarrow \left\{ \matrix{
\uparrow {H_2}:0,3 \hfill \cr
{\rm{dd}}\,A\buildrel { + NaOH\,du} \over
\longrightarrow \downarrow B\buildrel {to} \over
\longrightarrow ran:18g \hfill \cr} \right. \hfill \cr
\buildrel {CuC{l_2}:0,5} \over
\longrightarrow \left\{ \matrix{
ran \hfill \cr
{\rm{dd}}\,D\buildrel {{t_o}} \over
\longrightarrow muoi:65g \hfill \cr} \right. \hfill \cr} \right.\)

a.

M + 2HCl → MCl₂ + H₂↑

MO + 2HCl → MCl₂ + H₂O

MCl₂ + 2NaOH → M(OH)₂↓ + 2NaCl

M(OH)₂ → MO + H₂O

M + CuCl₂ → MCl₂ + Cu↓

b.

n_H2 = 0,3 → n_M = 0,3

\(\eqalign{
& \left\{ \matrix{
M:0,3 \hfill \cr
MO:x \hfill \cr
MC{l_2}:y \hfill \cr} \right.\buildrel {TGKL} \over
\longrightarrow \left\{ \matrix{
\underbrace {M \to MO}_{ \uparrow 16} \hfill \cr
\underbrace {MC{l_2} \to MO}_{ \uparrow 55} \hfill \cr} \right. \to - 16.0,3 + 55y = 18,7 - 18 = > y = 0,1 \cr
& {\rm{dd}}\,{\rm{D}}\underbrace {\left\{ \matrix{
CuC{l_2}\,du:0,2 \hfill \cr
MC{l_2}:0,4 \hfill \cr} \right.}_{65g} \to M = 24g \to \left\{ \matrix{
\% Mg = 38,5\% \hfill \cr
\% MgO = 10,07\% \hfill \cr
\% MgC{l_2} = 50,80\% \hfill \cr} \right. \cr} \)