Hòa tan hòa toàn a gam kim loại M (có hóa trị khôn...

- Hướng dẫn giải

Phương pháp giải:

bảo toàn khối lượng, bảo toàn nguyên tố.

Giải chi tiết:

\(\eqalign{
& M + HCl \to \left\{ \matrix{
\uparrow {H_2} \hfill \cr
{\rm{dd}}\,{\rm{D}}\left\{ \matrix{
MC{{\mathop{\rm l}\nolimits} _n}\buildrel { + NaHC{O_3}:0,2mol} \over
\longrightarrow \hfill \cr
HCl\,du \hfill \cr} \right. \hfill \cr} \right.\left\{ \matrix{
\uparrow C{O_2}:0,2 \hfill \cr
{\rm{dd}}\,E\left\{ \matrix{
NaCl:2,378\% \hfill \cr
MC{{\mathop{\rm l}\nolimits} _n}:7,724\% \hfill \cr} \right.\buildrel { + NaOH\,du} \over
\longrightarrow \downarrow \buildrel {to} \over
\longrightarrow {M_2}{O_n}:16g \hfill \cr} \right. \cr
& \buildrel {BT\,Na} \over
\longrightarrow \left\{ \matrix{
{n_{NaCl}} = {n_{NaHC{O_3}}} \hfill \cr
= > {n_{NaCl}} = 0,2 \hfill \cr} \right.\buildrel {\% NaCl:2,378\% } \over
\longrightarrow {\rm{dd}}\,E\buildrel {\% MC{l_n}:7,724\% } \over
\longrightarrow MC{{\mathop{\rm l}\nolimits} _n}:38g \cr
& \buildrel {BT\,M} \over
\longrightarrow \left\{ \matrix{
MC{{\mathop{\rm l}\nolimits} _n}:x \hfill \cr
{M_2}{O_n}:0,5x \hfill \cr} \right. \to \left\{ \matrix{
(M + 35,5n)x = 38 \hfill \cr
(2M + 16n).0,5x = 16 \hfill \cr} \right. \to \left\{ \matrix{
M.x = 9,6 \hfill \cr
n.x = 0,8 \hfill \cr} \right. = > M = 12n = > \left\{ \matrix{
n = 2 \hfill \cr
M = 24 \hfill \cr} \right. \cr
& \mathop {Mg}\limits_{0,4} {\rm{ }} + {\rm{ }}\mathop {2HCl}\limits_{0,8} {\rm{ }} \to {\rm{ }}MgC{l_2} + {\rm{ }}{H_2} \cr
& = > \,{n_{HCl\,bd}} = 1mol \cr} \)

BTKL: m_D + m_NaHCO3 = m_CO2 + m_E

m_D + 179,88 = 44.0,2 + 492 => m_D = 320,92

BTKL: m_Mg + m_ddHCl = m_H2  + m_D

=> 24 . 0,4 + m_ddHCl = 2 . 0,4 + 320,92 => m_ddHCl = 312,12

=> C%HCl = 11,69%