Chương 5. Nhóm Halogen

Refer

Tham Khảo

gọi a là số mol mgcl2

      blaf số mol kcl 

=> 95a +74,5b =3,93 (g)   (1)

cl-    +   ag+   =>     agcl

nag+ =0,07mol

=>ncl- =0,07 mol

=>2a +b =0,07 (2)

từ 1 và 2 ta có 

\(\left\{{}\begin{matrix}95a+74,5b=3,93\\2a+b=0,07\end{matrix}\right.\) =>\(\left\{{}\begin{matrix}a=\dfrac{257}{10800}\\b=\dfrac{121}{5400}\end{matrix}\right.\)

\(n_{O_2}=\dfrac{22,12-21,16}{32}=0,03\left(mol\right)\)

\(n_{KMnO_4}=\dfrac{22,12}{158}=0,14\left(mol\right)\)

Mn⁺⁷ + 5e --> Mn⁺²

0,14-->0,7

2O^-2 - 4e --> O₂⁰

         0,12<--0,03

2Cl^- - 2e --> Cl₂⁰

          2a<---a

Bảo toàn e: 2a + 0,12 = 0,7

=> a = 0,29 (mol)

=> V_Cl2 = 0,29.22,4 = 6,496 (l)

(Al,Fe)+HCl ->(AlCl3+FeCl2)+H2

khối lượng dd tăng thêm =7,8gam=mKL-mH2=8,3-mH2

=>mH2=8,3-7,8=0,5gam=>nH2=\(\dfrac{0,5}{2}\)=0,25 mol

bảo toàn H :nHcl=nH (trong h2)=2nH2=2.0,25=0,5mol

 nCl(trong muối)=nCl( trong HCl)=nHCl=0,5mol

mMuối =mKL+mCl=8,3+0,5.35,5=26,05gam

Ta có PTHH : 2KMnO4 + 16HCl → 2KCl + 2MnCl2 + 5Cl2 + 8H2O

Ta có: nKMnO4= \(\dfrac{31,6}{39+55+16.4}\)
=>nKMnO4= 0,2 (mol)
Theo PTHH nCl_{2  75%}=\(\dfrac{5}{2}\)nKMnO₄= \(\dfrac{5}{2}\).0,2= 0,5 (mol)

Biết Hiệu suất của pứ là 75% 
=> nCl_{2=\(\dfrac{0,5.75}{100}\)}=0,375 (mol)
=>VCl₂=0,375.22,4=8,4 (l)

 

a) 
\(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)

\(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)

\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3\downarrow+3NaCl\)

\(2NaCl+H_2SO_4\underrightarrow{t^o}Na_2SO_4+2HCl\)

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

\(CuCl_2+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2AgCl\downarrow\)

b)

\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)

\(H_2+Cl_2\underrightarrow{t^o}2HCl\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(FeCl_3+3AgNO_3\rightarrow Fe\left(NO_3\right)_3+3AgCl\downarrow\)

\(2AgCl\underrightarrow{as}2Ag+Cl_2\)

\(2NaBr+Cl_2\rightarrow2NaCl+Br_2\)

\(2NaI+Br_2\rightarrow2NaBr+I_2\)

\(Zn+I_2\underrightarrow{H_2O}ZnI_2\)

\(ZnI_2+2NaOH\rightarrow2NaI+Zn\left(OH\right)_2\)

c)
\(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)

\(3Cl_2+6KOH\underrightarrow{t^o}5KCl+KClO_3+3H_2O\)

\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

\(2KCl+H_2SO_4\underrightarrow{t^o}K_2SO_4+2HCl\)

\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)

\(Ca\left(OH\right)_2+Cl_2\rightarrow CaOCl_2+H_2O\)

nKOH = 0,2 . 1 = 0,2 (mol)

PTHH: 2KOH + Cl2 -> KCl + KClO + H2O

nKCl = nKClO = 0,2/2 = 0,1 (mol)

CMKCl = CMKClO = 0,1/0,2 = 0,5M

=> A