Hỗn hợp A gồm MgO và CaO, hỗn hợp B gồm MgO và Al

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

\(\eqalign{
& \underbrace {A\left\{ \matrix{
MgO:x \hfill \cr
CaO:y \hfill \cr} \right.}_{9,6g}\buildrel { + HCl:0,57} \over
\longrightarrow {\rm{dd}}\,X\buildrel { + N{a_2}C{O_3}} \over
\longrightarrow \underbrace { \uparrow C{O_2}}_{0,085} \cr
& a. \cr
& Pt:{\rm{ }} \cr
& 2HCl{\rm{ }} + {\rm{ }}N{a_2}C{O_3} \to {\rm{ }}2NaCl{\rm{ }} + {\rm{ }}C{O_2} \uparrow {\rm{ }} + {\rm{ }}{H_2}O \cr
& 0,17{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{ }}\,\,\,\,\,\,\,\,\,\,\, \leftarrow 0,085 \cr
& \to {\rm{ }}{n_{HClpu}} = {\rm{ }}0,4 \cr
& MgO{\rm{ }} + {\rm{ }}2HCl{\rm{ }} \to {\rm{ }}MgC{l_2} + {\rm{ }}{H_2}O \cr
& CaO{\rm{ }} + {\rm{ }}2HCl{\rm{ }} \to {\rm{ }}CaC{l_2} + {\rm{ }}{H_2}O \cr
& \to \left\{ \matrix{
40x + 56y = 9,6 \hfill \cr
2x + 2y = 0,4 \hfill \cr} \right. \to \left\{ \matrix{
x = 0,1 \hfill \cr
y = 0,1 \hfill \cr} \right. \to \left\{ \matrix{
\% MgO = 25\% \hfill \cr
\% CaO = 75\% \hfill \cr} \right. \cr
& b. \cr
& {m_X}{\rm{ }} = {\rm{ }}{m_A}{\rm{ }} + {\rm{ }}{m_{ddHCl}}{\rm{ }} = {\rm{ }}114,3{\rm{ }}\left( g \right) \cr
& A\left\{ \matrix{
MgO:0,1 \hfill \cr
CaO \hfill \cr} \right.\buildrel {{m_{MgO(A)}} = 1,125\,.\,{m_{MgO(B)}}} \over
\longrightarrow B\left\{ \matrix{
MgO:0,1125 \hfill \cr
A{l_2}{O_3}:0,05 \hfill \cr} \right. \to Y\left\{ \matrix{
\buildrel {BTNT\,Mg} \over
\longrightarrow MgC{l_2}:0,1125 \hfill \cr
\buildrel {BTNT\,Al} \over
\longrightarrow AlC{l_3}:0,1 \hfill \cr
\buildrel {BTNT\,Cl} \over
\longrightarrow HCl{\,_{du}}:0,045 \hfill \cr} \right. \cr
& Y\left\{ \matrix{
MgC{l_2}:0,1125 \hfill \cr
AlC{l_3}:0,1 \hfill \cr
HCl{\,_{du}}:0,045 \hfill \cr} \right.\buildrel { + KOH:0,68mol} \over
\longrightarrow {\rm{dd}}Y\left\{ \matrix{
\buildrel {BTNT\,Cl} \over
\longrightarrow KCl:0,57 \hfill \cr
\buildrel {BTNT\,Al} \over
\longrightarrow KAl{O_2}:0,1 \hfill \cr
\buildrel {BTNT\,K} \over
\longrightarrow KO{H_{du}}:0,01 \hfill \cr} \right. = > {m_{Al{{(OH)}_3} \downarrow }} = 0g \cr} \)