1. Cho m gam hỗn hợp A (dạng bột)...

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

Hướng dẫn giải:

\(\underbrace {A\left\{ \matrix{
Fe:x \hfill \cr
Cu:y \hfill \cr} \right.}_{m(g)}\buildrel { + AgN{O_3}:0,1mol} \over
\longrightarrow \left\{ \matrix{
{m_r} = 13,36g(2\,KL) \hfill \cr
{\rm{dd}}\,Y\buildrel { + NaOH\,du} \over
\longrightarrow \downarrow Z\buildrel {{t^o}} \over
\longrightarrow p(g) \hfill \cr} \right.\)

Rắn phải có Ag, mà có 2 kim loại nên kim loại còn lại là Cu. Kim loại dư nên AgNO₃hết

\(\eqalign{
& \underbrace {ran}_{13,36g}\left\{ \matrix{
\buildrel {BTNT\,Ag} \over
\longrightarrow Ag:0,1 \hfill \cr
Cu:0,04 \hfill \cr} \right. \to {\rm{dd}}\,Y\underbrace {\left\{ \matrix{
\buildrel {BTNT\,Fe} \over
\longrightarrow Fe{(N{O_3})_2} \hfill \cr
\buildrel {BTNT\,Cu} \over
\longrightarrow Cu{(N{O_3})_2}:y - 0,04 \hfill \cr} \right.}_{\buildrel {BT\,NO_3^ - } \over
\longrightarrow 2x + 2(y - 0,04) = 0,1(*)} \to ran\left\{ \matrix{
F{e_2}{O_3}:0,5x \hfill \cr
CuO:y - 0,04 \hfill \cr} \right. \cr
& = > p = 160\,.\,0,5x + 80(y - 0,04) = > p = 80(x + y) - 3,2\buildrel {(*)} \over
\longrightarrow p = 4g \cr
& b. \cr
& \left\{ \matrix{
2x + 2(y - 0,04) = 0,1 \hfill \cr
56x + 64y = 5,44 \hfill \cr} \right. \to \left\{ \matrix{
x = 0,04 \hfill \cr
y = 0,05 \hfill \cr} \right. \to {m_{Cu}} = 3,2g \cr} \)

2.

a. Ta có:

\(\eqalign{
& {{2X} \over {16\,.\,7}} = {{71} \over {112}} = > X = 35,5 = > X:Cl \cr
& b. \cr
& \underbrace {\left\{ \matrix{
Al \hfill \cr
MgC{O_3} \hfill \cr} \right.}_{4,32g}\buildrel { + HCl:\,0,3mol} \over
\longrightarrow \left\{ \matrix{
\underbrace { \uparrow ({H_2},C{O_2})}_{\overline M = 27,2} \hfill \cr
{\rm{dd}}\,A \hfill \cr} \right. \cr
& Al{\rm{ }} + {\rm{ }}3HCl{\rm{ }} \to {\rm{ }}AlC{l_3} + {\rm{ }}1,5{H_2} \uparrow \cr
& MgC{O_3} + {\rm{ }}2HCl{\rm{ }} \to {\rm{ }}MgC{l_2} + {\rm{ }}C{O_2} \uparrow {\rm{ }} + {\rm{ }}{H_2}O \cr
& \left\{ \matrix{
Al:a \hfill \cr
MgC{O_3}:b \hfill \cr} \right. \to \left\{ \matrix{
27a + 84b = 4,32 \hfill \cr
2\,.\,1,5a + 44b = 27,2(a + b) \hfill \cr} \right. \to \left\{ \matrix{
a = 0,02 \hfill \cr
b = 0,045 \hfill \cr} \right. \cr
& = > {n_{HCl\,pu}} = {\rm{ }}2.{n_{{H_2}}} + {\rm{ }}2.{n_{C{O_2}}} = {\rm{ }}0,13{\rm{ }} \to {\rm{ }}{n_{HCl\,du}} = {\rm{ }}0,17 \cr
& \buildrel {BTKL} \over
\longrightarrow \left\{ \matrix{
{m_{KL}} + {m_{{\rm{dd}}}} = {m_ \uparrow } + {m_{{\rm{dd}}\,A}} \hfill \cr
\to 4,32 + 200.1,05 = 2.0,03 + 44.0,045 + {m_{{\rm{dd}}\,A}} \hfill \cr} \right. \to {m_{{\rm{dd}}\,A}} = 212,28g \cr
& \to \left\{ \matrix{
C\% AlC{l_3} = 1,26\% \hfill \cr
C\% MgC{l_2} = 2,01\% \hfill \cr
C\% HCl\,du = 2,92\% \hfill \cr} \right. \cr} \)