Bài 22: Clo

\(2KMnO_4+16HCl_{đặc,nóng}\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\\ MnO_2+4HCl_{đặc,nóng}\rightarrow MnCl_2+Cl_2+2H_2O\\ Đặt:n_{KMnO_4}=a\left(mol\right);n_{MnO_2}=b\left(mol\right)\left(a,b>0\right)\\ Vì:n_{Cl_2}=\dfrac{9,632}{22,4}=0,43\left(mol\right)\\ \Rightarrow2,5a+b=0,43\left(1\right)\\ Ta.có:n_O=4a+2b\Rightarrow m_O=16.\left(4a+2b\right)=64a+32b=0,39114.\left(158a+87b\right)\\ \Leftrightarrow64a+32b-61,80012a-34,02918b=0\\ \Leftrightarrow2,19988a-2,02918b=0\left(2\right)\\ \left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}2,19988a-2,02918b=0\\2,5a+b=0,43\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,12\\b=0,13\end{matrix}\right.\)

\(n_{MnCl_2}=n_{MnO_2}=a+b=0,25\left(mol\right)\\ \Rightarrow m_{MnCl_2}=126.0,25=31,5\left(g\right)\)

=>Chọn D

(1): \(2HCl\rightarrow H_2+Cl_2\)

(2): \(3Cl_2+2Fe\rightarrow2FeCl_3\)

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

(4): \(NaCl+H_2SO_4\rightarrow HCl+NaHSO_4\)

(5): \(2HCl+CuSO_4\rightarrow CuCl_2+H_2SO_4\)

(6): \(CuCl_2+2AgNO_3\rightarrow2AgCl+Cu\left(NO_3\right)_2\)

\(MnO_2+4HCl_{đặc,nóng}\rightarrow MnCl_2+Cl_2+2H_2O\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3\downarrow+3NaCl\\ NaCl+H_2SO_{4\left(đặc\right)}\left(250^oC\right)NaHSO_4+HCl\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ CuCl_2+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2AgCl\downarrow\)

Cl₂+2NaBr ->2NaCl+Br₂

  0,04 -0,08-------0,08-----0,04 mol

Cl₂+2NaI ->2NaCl+I₂

0,03--0,06---0,06-----0,03 mol

n _NaBr =0,8.0,1=0,08 mol

n _NaI=0,6.0,1=0,06 mol

=>V _Cl2=(0,04+0,03).22,4=1,568l

=>m _Br2=0,04 .160=6,4g

=>m _I2=0,03.254=7,62g

b)

NaCl+AgNO₃->AgCl+NaNO₃

0,14-----------------0,14 mol

=>m _AgCl=0,14.143,5=20,09g

c)2KMnO₄+16HCl->2KCl+2MnCl₂+5Cl₂+8H₂O

    0,028--------------------------------------0,07 mol

=>m _KMnO4=0,028.158=4,424g

$Cl_2 + H_2O \rightleftharpoons HCl  + HClO \\ Cl_2 + 2NaOH \to NaCl + NaClO + H_2O\\ MnO_2 + 4HCl \xrightarrow{t^o} MnCl_2 + Cl_2 + 2H_2O \\ 2KMnO_4 + 16HCl \to 2MnCl_2 + 2KCl + 5Cl_2 + 8H_2O \\ 2NaCl + 2H_2O \xrightarrow{điện\ phân\ dung\ dịch,có\ màng\ ngăn} 2NaOH + H_2 + Cl_2$

\(2Cl_2+2H_2O\rightarrow4HCl+O_2\)

\(2Cl_2+4NaOH\rightarrow4NaCl+2H_2O\)

\(2NaCl+2H_2O\rightarrow2NaOH+Cl_2+H_2\)

\(Cl_2+H_2O⇌HCl+HClO\)

\(Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O\)

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

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

\(2NaCl+2H_2O\underrightarrow{đpcmn}2NaOH+H_2+Cl_2\)

\(2Na+Cl_2\underrightarrow{t^o}2NaCl\)

\(2Al+3Cl_2\underrightarrow{t^o}2AlCl_3\)

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

\(Cu+Cl_2\underrightarrow{t^o}CuCl_2\)

\(Cl_2+H_2\underrightarrow{as}2HCl\)

$2Na + Cl_2 \xrightarrow{t^o} 2NaCl$
$2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3$

$2Fe +3Cl_2 \xrightarrow{t^o} 2FeCl_3$
$Cu + Cl_2 \xrightarrow{t^o} CuCl_2$
$H_2 + Cl_2 \xrightarrow{ánh\ sáng} 2HCl$

chj kb vs em vs ạ em gửi tin nhắn rùi:)

\(n_{Cl_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\)

PTHH: 2M + Cl₂ --t^o--> 2MCl

________0,075---->0,15

=> \(M_{MCl}=\dfrac{11,175}{0,15}=74,5\left(g/mol\right)\)

=> M_M = 39 (K)

\(Zn+X_2-^{t^o}\rightarrow ZnX_2\\ Tacó:n_{Zn}=n_{ZnX_2}\\ \Rightarrow\dfrac{9,75}{65}=\dfrac{20,4}{65+2X}\\ \Rightarrow X=35,5\left(Cl\right)\)

a) nCl2=7,28/22,4=0,325(mol)

=> mCl2=0,325.71=23,075(mol)

=> m(muối)= m(hh)+ mCl2= 10,45+23,075=33,525(g)

b) PTHH: 2 Al + 3 Cl2 -to-> 2 AlCl3

a__________1,5a(mol)

Cu + Cl2 -to-> CuCl2

b___b(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}27a+64b=10,45\\1,5a+b=0,325\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,15\\b=0,1\end{matrix}\right.\)

=> mCu=0,1.64=6,4(g)

=>%mCu= (6,4/10,45).100=61,244%

=>%mAl=38,756%

Đặt : nKMnO4 = 3a(mol) 

nKClO3 = 4a (mol) 

mX = 3a * 158 + 4a * 122.5 = 48.2 (g) 

=> a = 0.05 

BTKL : 

mO2 = 48.2 - 40.2 = 8 (g) 

nO2 = 8/32 = 0.25 (mol) 

BT e : 

5nKMnO4 + 6nKClO3 = 4nO2 + 2nCl2 

=> nCl2 = 5*0.15 + 6*0.2 - 4*0.25 = 0.95 (mol) 

VCl2 = 21.28 (l) 

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