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Study of aluminium chloride - class-XII

Description: study of aluminium chloride
Number of Questions: 40
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Tags: some p-block elements elements of group 13, 14 and 15 chemistry p- block elements-i
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Some properties of aluminium are listed.
1 It has mechanical strength.
2 It is resistant to corrosion.
3 It has a low density.
4 It conducts heat.
Which three properties make aluminium useful for making the bodies of aircraft?

  1. $1, 2$ and $3$

  2. $1, 2$ and $4$

  3. $1, 3$ and $4$

  4. $2, 3$ and $4$


Correct Option: A
Explanation:

Solution:- (A) $1, \; 2$ and $3$

Aluminium is useful in making of aircraft bodies due to its mechanical strength, low density and corrosion resistance.

Element "x" which is solid and having high melting point, form a Chloride $"x  Cl 3"$. This element "x" would be in which group of Periodic Table__.

  1. Na

  2. Mg

  3. Al

  4. Si


Correct Option: C
Explanation:
Element aluminium has atomic number $13$. Electronic configuration is $2, 8, 3$ 
Valency $=3$
$2Al+3Cl _2\rightarrow 2AlCl _3$

Anhydrous $AlCl _{3}$ is formed in:

  1. $AlCl _{3}.6H _{2}O \xrightarrow[]{\Delta}$

  2. $Al(red hot) + HCl(moist)\xrightarrow[]{\Delta}$

  3. $Al _{2}O _{3} + C+ Cl _{2}(dry) \xrightarrow[]{\Delta}$

  4. All of these


Correct Option: A

Element 'x' which is solid and having high melting point, form a Chloride $'x Cl _3'$. This element 'x' would be in which group of Periodic table -

  1. Na

  2. Mg

  3. Al

  4. Si


Correct Option: C
Explanation:

Element aluminium has atomic number 13. Electronic configuration is 2, 8, 3
$Valency = 3$
$2Al + 3Cl _2 \rightarrow 2AlCl _3$

Which of the following species do not exist?

  1. $Al _2Cl _6$

  2. $BH _3$

  3. $PH _3$

  4. $PCl _5$


Correct Option: B
Explanation:
  • $BH _3$ molecule is electron deficient in nature since 'B' atom has only 6 electrons around it. So it exists in the form of dimer that is $B _2H _6.$
  • Hence option B is correct answer.

On the hydrolysis of $Al{Cl} _{3}$ then the no. of ${Al}^{+3}$ will be:

  1. Is equal to chloride ions

  2. Is equal to $\cfrac{1}{3}$ chloride ions

  3. Three times of ${Cl}^{-}$ ion

  4. None of these


Correct Option: B
Explanation:

The hydrolysis reaction will be $AlC{ l } _{ 3 }+{ 3H } _{ 2 }O\rightarrow Al(OH{ ) } _{ 3 }+3HCl$

So the number of $A{ l }^{ +3 }$ will be equal to $\frac { 1 }{ 3 } $ chloride ions.

Anhydrous $Al{Cl} _{3}$ cannot be obtained from which of the following reactions :

  1. Heating $Al{Cl} _{3}.6{H} _{2}O$

  2. By passing dry $HCl$ over hot aluminium powder

  3. By passing dry ${Cl} _{2}$ over hot aluminium powder

  4. By passing dry ${Cl} _{2}$ over hot mixture of alumina and coke


Correct Option: A
Explanation:

On heating $AlCl _{ 3 }.6{ H } _{ 2 }O$ hydrolysis takes place to produce $Al\left( OH \right) _{ 3 }$ and $HCl$
and on further heating $Al\left( OH \right) _{ 3 }$ converts to $Al _{ 2 }O _{ 3 }$.

By passing dry $HCl$ or ${ Cl } _{ 2 }$ over heated aluminium,
$2Al+3C{ l } _{ 2 }\longrightarrow 2AlC{ l } _{ 3 }$

$2Al+6HC{ l }\longrightarrow 2AlC{ l } _{ 3 }+3{ H } _{ 2 }$

Macah's method:$Al _{ 2 }O _{ 3 }+3C+3{ Cl } _{ 2 }(dry)\xrightarrow [ \Delta  ]{  } 2AlC{ l } _{ 3 }+3CO\uparrow $

Is ever $AlCl _4$ possible ?

  1. True

  2. False


Correct Option: A
Explanation:
  • There is dative coordination from a lone pair on a chloride ion to the vacant orbital on the aluminium ion. 
  • This gives the aluminium ion a full octet  and does not change the valency rules of aluminium.
  • Hence $AlCl _4$ is possible.

Which of the following statements is not correct about $AlCl _3$ ?

  1. It exists as a dimer

  2. It is a covalent compound

  3. It involves back bonding between Cl and Al

  4. Its aqueous solution conducts electricity


Correct Option: C
Explanation:

Due to high size of chlorine ,there is no back bonding in$AlCl _3$

so, option c is incorrect 

$AlF _{ 3 }$ is soluble in HF only in presence of KF. its is due to the formation of:

  1. $AlH _{ 3 }$

  2. $K\left[ AlF _{ 3 }H \right]$

  3. ${ K } _{ 3 }\left[ AlF _{ 3 }{ H } _{ 3 } \right]$

  4. ${ K } _{ 3 }\left[ AlF _{ 6 } \right]$


Correct Option: D
Explanation:

$AlF _3$ is insoluble in the anhydrous $HF$ because the $F$ ions are not available in intermolecular hydrogen bonded $HF$ but it becomes soluble in the  presence of $KF$ due to the formation of soluble complex,

$K _3[AlF _6]$
$AlF _3+3KF\longrightarrow K _3[AlF _6]$

$AlCl _3+NH _4OH\rightarrow X$ ; x is 

  1. A white gelatinous precipitate

  2. Soluble in excess of $NH _4OH$

  3. Soluble in excess of $NaOH$

  4. Amphoteric in nature


Correct Option: A

 ${ AlF } _{ 3 }$ is soluble in $HF$ only in presence of $KF$. It is due to the formation of:

  1. ${ AlH } _{ 3 }$

  2. ${ K[AlF } _{ 3 }H]$

  3. ${ { K } _{ 3 }[AlF } _{ 3 }{ H } _{ 3 }]$

  4. ${ { K } _{ 3 }[AlF } _{ 6 }]$


Correct Option: A
Explanation:
$AlF _3$  is insoluble in the anhydrous $HF$ because the $F^-$ ions are not available in intermolecular hydrogen bonded HF, but it becomes soluble in the presence of $KF$ due to the formation of soluble complex, $K _3[AlF _6].$
$AlF _3+3KF⟶K _3[AlF _6]$
Hence option D is correct answer.

Which have Vont Hoff factor same as $K _4 [Fe(CN) _6]$ 

  1. $Al _2(SO _4) _3$

  2. $Mg(NO _3) _2$

  3. $CaCl _2$

  4. $NaNO _3$


Correct Option: A
Explanation:

$i$ for $K _4 [Fe(CN) _6]$ is $5$ and equal to $i$ of $Al _2(SO _4) _3$ 

The molecule which is NOT hydrolysed by water at $25^oC$ is

  1. $AICI _3$

  2. $SiCI _4$

  3. $BF _3$

  4. $SF _6$


Correct Option: A

${Al} _{2}{O} _{3}$ is reduced by electrolysis at low potential and high currents. If $4.0\times {10}^{4}$ amperes of current is passed through molten ${Al} _{2}{O} _{3}$ for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency, At. mass of $Al=27g{mol}^{-1}$)

  1. $1.3\times {10}^{4}g$

  2. $9.0\times {10}^{3}g$

  3. $8.1\times {10}^{4}g$

  4. $2.4\times {10}^{5}g$


Correct Option: C
Explanation:

$Al _2O _3(l) \rightarrow2Al^{3+}(l) + 3O^{2-}(l) $

Reduction equation of $Al^{3+}: $
$Al^{3+}(aq) + 3e^- \rightarrow Al(s) $

As,

Ampere = Coulomb / sec 

Time unit should be second. 
$6 hr \times 60 \dfrac{min}{ hr} \times 60 \dfrac{sec}{ min} = 21600 sec $
Coulomb = Ampere x second $= 40000 \times 21600 = 8.64\times10^8 $

1 mol $e^-$ carries a charge of 1 faraday, or 96485 coulombs. 
Then the mole of $e^-$ used in the electrolysis will be: 

$\dfrac{8.64\times10^8   \text{coulombs}}{ 96485 -\text{coulombs/mol}}- of -e^- = 8954.8  \text{mol of} -e^- $

According to the reduction equation, 3 moles of e- are needed to reduce 1 mol of Al^3+ to Al metal. 
Then , the mole of Al to be deposited at the cathode will be; 
$\dfrac{8954.8 \text{mol e-} }{ \dfrac{3 -\text{mol of e-}}{\text{mol Al}}} = 2985\text{ mol Al} $

Molar mass of $Al : 26.98 g/mol $
Mass of Al produced$: 2985 mol \times 26.98 g/mol = 80535 g $
$Al = 8.1\times10^4 g$

Alluminium chloride nexists as dimer, $A{l _2}C{l _6}$ in solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives:

  1. ${{\text{[Al(OH}}{{\text{)}} _{\text{6}}}{\text{]}}^{{\text{3 - }}}}{\text{ + 3HCl}}$

  2. ${{\text{[Al(}}{{\text{H}} _2}{\text{O}}{{\text{)}} _{\text{6}}}{\text{]}}^{{\text{3 + }}}}{\text{ + 3HC}}{{\text{l}}^ - }$

  3. $A{l^{3 + }} + 3C{l^ - }$

  4. ${\text{A}}{{\text{l}} _{\text{2}}}{{\text{O}} _{\text{3}}}{\text{ + 6HCl}}$


Correct Option: B

Aluminium chloride exists as dimer, $Al _2Cl _6$, in solid state as well as in solution of non-polar solvents such as $C _6H _6$.


 When dissolved in water it gives :

  1. $Al _2O _3+6HCl$

  2. $[Al(H _2O) _6]^{3+}+3Cl^-$

  3. $[Al(OH) _6]^{3+}+6Cl^-$

  4. $Al^{3+}+3Cl^-$


Correct Option: B
Explanation:

Reaction:

$AlCl _3 $ is covalent but in water it becomes ionic due to large hydration energy oh $Al^{3+}$.

$AlCl _3 + 6H _2O \rightarrow [Al(H _2O) _6]^{3+} + 3Cl^-$.


Hence,option B is correct.

$AlCl _3$ on hydrolysis gives :

  1. $Al _2O _3\cdot H _2O$

  2. $Al(OH) _3$

  3. $Al _2O _3$

  4. $AlCl _3\cdot 6H _2O$


Correct Option: B
Explanation:

$AlCl _3$ on hydrolysis gives $Al(OH) _3$

Hydrolysis is breaking with the help of water.

$AlCl _3 + 3 H _2O\rightarrow Al(OH) _3+ 3 HCl$

Hence option B is correct.

Aqueous solution of $AIC{I _3}$ on heating to dryness will give 

  1. $AIC{I _3}$

  2. $A{l _2}{O _3}$

  3. $Al(OH)C{L _2}\,$

  4. $A{L _2}C{l _6}$


Correct Option: B

Aluminium (III) chloride forms a dimer because aluminium :

  1. cannot form a trimer.

  2. has high ionization energy.

  3. belongs to third group.

  4. can have higher coordination number.


Correct Option: D
Explanation:

 Aluminium trichloride is a dimer, $Al _2Cl _6$.

This is because: 
$AlCl _3$ forms a dimer as there is unsaturation in coordination. Also it enables Al atoms to complete their octets.

Aluminium is more like a metalloid than a metal.The electronegativity of aluminium is higher than the electronegativity of sodium .


$AlCl _{3}$ on hydrolysis gives :

  1. $Al _{2}O _{3}.H _{2}O$

  2. $Al(OH) _{3}$

  3. $Al _{2}O _{3}$

  4. $AlCl _{3}.6H _{2}O$


Correct Option: B
Explanation:

$AlCl _{3}$ on hydrolysis gives $Al(OH) _{3}$
$AlCl _{3}+3H _{2}O\rightarrow Al(OH) _{3}+3HCl$
Hence option B is correct.

$AlX _3$ (X = Cl, Br) exists as dimer and retains dimer formula in non-polar solvents like ether and benzene.

  1. True

  2. False


Correct Option: A
Explanation:

In the liquid state aluminium trichloride exists as the dimer with tetracoordinate aluminium. This change in structure is related to the lower density of liquid phase. These dimers are also found in the vapor phase. At higher temperature dimers dissociate into trigonal planar. The melt conducts electricity poorly.

Which reaction cannot give anhydrous $AlCl _3$ ?

  1. Heating of $AlCl _3.\, 6H _2O$

  2. Passing dry HCl over heated aluminium powder

  3. Passing dry $Cl _2$ over heated aluminium powder

  4. Heating a mixture of alumina and coke in a current of dry $Cl _2$


Correct Option: A
Explanation:

Aluminium chloride reacts with water to form  hexahydrate. The anhydrous phase cannot be regained on heating as HCl is lost leaving aluminium hydroxide or alumina.
$Al\left( { H } _{ 2 }O \right) _{ 6 }{ Cl } _{ 3 }\rightarrow { Al\left( OH \right)  } _{ 3 }+3HCl+{ 3H } _{ 2 }O$

Which of the following statements is correct ?

  1. $BCl _{3}$ and $AlCl _{3}$ are both Lewis acids and $BCl _{3}$ is stronger than $AlCl _{3}$

  2. $BCl _{3}$ and $AlCl _{3}$ both Lewis acids and $AlCl _{3}$ is stronger than $BCl _{3}$

  3. Both $BCl _{3}$ and $AlCl _{3}$ are both equally strong Lewis acids.

  4. Both $BCl _{3}$ and $AlCl _{3}$ are not Lewis acids.


Correct Option: A

$AlCl _3$ is:

  1. anhydrous and covalent

  2. anhydrous and ionic

  3. covalent and basic

  4. coordinate and acidic


Correct Option: A

Alumina is:

  1. acidic

  2. basic

  3. amphoteric

  4. none of the above


Correct Option: C
Explanation:

Allumina is $Al _2O _3$ reacts as both acid and base in chemical reactions. So it is called amphoteric oxide.
Hence option $C$ is correct.

Anhydrous $AlCl _3$ is prepared by :

  1. reaction of HCl and Al metal

  2. reaction of dry HCl and heated Al metal

  3. passing $Cl _2$ gas over heated Al metal

  4. reaction of $Al _2O _3$ with dil HCl.


Correct Option: B
Explanation:
Answer:- (B) reaction of dry $HCl$ and heated $Al$ metal.
Aluminium chloride is manufactured on a large scale by the exothermic reaction of heated aluminium metal with dry hydrogen chloride at temperatures between $650$ to $750 ^0C$
$2Al + 6HCl \; \longrightarrow \; 2Al{Cl} _3 + 3{H} _2$

$AlCl _3$ achieves stability by forming a dimer. In trivalent state the compound is hydrolysed in water. $AlCl _3$ in acidified aqueous solution forms :

  1. $Al(OH) _3+HCl$

  2. $[Al(H _2O) _6]^{3+}+3Cl^-$

  3. $AlCl _3.2H _2O$

  4. $Al _2O _3+HCl$


Correct Option: B
Explanation:
Answer:- (B) $[Al({H _2O)} _6]^{+3} +3{Cl}^{-}$

The electron deficiency of Aluminium atom in $Al{Cl} _3$ is compensated by formation of co-ordinate bond between lone pair of Chlorine atom of another $Al{Cl} _3$ molecule and the empty unhybridised p orbital of Aluminium atom, thus forms dimer and in acidified aqueous solution forms $[Al({H _2O)} _6]^{+3} +3{Cl}^{-}$.

$Al{Cl} _3 + 6H _2O \; \longrightarrow \; [Al({H _2O)} _6]^{+3} + 3{Cl}^{-}$

Aluminium chloride exists as dime , $Al _2Cl _6$ in solid state as well as in solution of non-polar solvents like benzene.When dissolved in water, it gives :

  1. $Al^{3+} +2Cl$

  2. $Al _2O _3+6HCl$

  3. $[Al(OH) _6]^3+3HCl$

  4. $[Al(H _2O) _6]^{3+}+3Cl$


Correct Option: D

Which of the following will be formed, if we heat an aqueous solution of $AlCl _3$ to dryness?

  1. Solid $AlCl _3$

  2. Dimer $Al _2Cl _6$

  3. $Al(OH) _3$

  4. $Al _2O _3$


Correct Option: D
Explanation:

We knoe $AlCl _3$ exists as a dimer $Al _2Cl _6$

The reactions involved are given by,
$Al _2Cl _6 .6H _2O \rightarrow Al _2O _3+3H _2O+6HCl$
So,option D is correct answer.

Anhydrous $AlCl _3$ fumes in air. What is the reason for it?

  1. It is hygroscopic in nature

  2. It gives out chlorine when exposed to air.

  3. It is hydrolysed in moist air giving out fumes of $HCl$.

  4. It loses water when exposed to moist air.


Correct Option: C
Explanation:
White fumes are due to the presence of $HCl$ gas.

Anhydrous aluminium chloride is hydrolysed partially with the moisture in the atmosphere to give $HCl$ gas. 

This HCl combines with the moisture in the air and appears white in colour.

$Al{Cl} _3 + 3H _2O \; \longrightarrow \; Al{(OH)} _3 + 3HCl \text{(white fumes)}$

Hence, the correct answer is option $\text{C}$.

Which reaction cannot give anhydrous $AlCl _{3}$

  1. Heating of $AlCl _{3}.6H _{2}O$

  2. Passing dry $HCl$ over heated aluminium powder

  3. Passing dry $Cl _{2}$ over heated aluminium powder

  4. Heating a mixture of alumina and coke in a current of dry $Cl _{2}$


Correct Option: A
Explanation:
Option A is correct because heating of $AlCl _{3}.6H _{2}O$ do not produce anhydrous $AlCl _{3}$ on heating

$AlCl _{3}.6H _{2}O\rightarrow Al _{2}O _{3}+HCl+H _{2}O$

Which of the following is a Lewis acid?

  1. $AlCl _3$

  2. $MgCl _2$

  3. $CaCl _2$

  4. $BaCl _2$


Correct Option: A
Explanation:
$Al$ needs two more electrons to complete its octet, it has the tendency to accept a pair of electrons to achieve stable electronic configuration. By definition, those which accepts electrons are called lewis acids. 

Thus, $Al{Cl} _{3}$ is a Lewis acid.


Hence, the correct answer is option $\text{A}$.

Which of the following statement about anhydrous aluminium chloride is correct?

  1. Its anhydrous form is deliquescent and fumes in air.

  2. It is not easily hydrolyzed.

  3. It sublimes at  $20^0C$ under vaccum.

  4. It is a strong Lewis base.


Correct Option: A
Explanation:

Aluminium chloride is hygroscopic, having a very pronounced affinity for water. It fumes in moist air and hisses when mixed with liquid water as the Cl ions are displaced with $H _2O$ molecules in the lattice to form the hexahydrate $AlCl _36H _2O$ (also white to yellowish in color).

It is  easily hydrolyzed.

It do not sublimes at $20^0C$ under vacuum.

It is a strong Lewis acid.(it accepts electrons)

So option A only is correct

Hence option A is correct statement.

The solubility of anhydrous $AlCl _{3}$ and hydrous $AlCl _{3}$ in diethyl ether are $S _{1}$ and $S _{2}$ respectively. Then -

  1. $S _{1}=S _{2}$

  2. $S _{1}> S _{2}$

  3. $S _{1}< S _{2}$

  4. $S _{1}< S _{2}$ but not $S _{1}=S _{2}$


Correct Option: B
Explanation:

Anhydrous AlCl3 will dissolve the diethyl ether because anhydrous ​AlCl3 is a good lewis acid. It is electron deficient. The lone pairs of diethyl ether will be donated to the anhydrous ​AlCl3 due to which it is soluble. On the other hand, hydrous ​AlCl3 have the water molecules which make it poor lewis acid.


So, 
$S _{1}> S _{2}$.

Which of the following methods can be used for the preparation of anhydrous aluminium chloride?

  1. Heating ${Al Cl _3.6 H _2 O}$

  2. Heating a mixture of alumina and coke in a current of dry chlorine

  3. Passing dry HCl gas over heated aluminium powder

  4. Passing dry chlorine over heated aluminium


Correct Option: B,C,D
Explanation:

Anhydrous aluminium chloride can be prepared by the following methods:
(B) Heating a mixture of alumina and coke in a current of dry chlorine.
$Al _2O _3+3C+3Cl _2 \rightarrow 2AlCl _3+3CO\uparrow$


(C) Passing dry HCl gas over heated aluminium powder.
$2Al+6HCl \rightarrow 2AlCl _3+3H _2$

(D) Passing dry chlorine over heated aluminium.
$2Al+3Cl _2 \rightarrow 2AlCl _3$

It can not be obtained by heating of $AlCl _3.6H _2O$, because $AlCl _3$ is covalent and hence when the water of crystallisation is evaporated, $AlCl _3$ will react with the water to form the hydroxide.

Which one of the following processes does not yield anhydrous $ AlCl _3 $?

  1. Heating of $ AlCl _3 . 6H _2O $.

  2. Passing dry HCl over heated aluminium powder.

  3. Passing dry $ Cl _2 $ over heated aluminium powder.

  4. Heating a mixture of alumina and coke in a current of dry $ Cl _2 $.


Correct Option: A
Explanation:

$ 2AlCl _3 . 6H _2O \xrightarrow {\Delta} Al _2O _3 + 6HCl + 9H _2O $ .

Anhydrous $AlCl _3$ is obtained from:

  1. dilute $HCl$ and $Al$

  2. $Al$ and $Cl$ gas

  3. $HCl$ gas and $Al$ metal

  4. none


Correct Option: C
Explanation:

When $Al$ metal is reacted with acid $HCl$ it libetrates hydrogen gas and  $AlCl _3$
$6HCl+2Al\rightarrow 2AlCl _3+3H _2$
Hence option $C$ is correct.

An inorganic compound $A$ shows the following reaction:
(i) It is white solid, exists as dimer and fumes in the wet air.
(ii)It sublimes at $180^{ \circ  }C$ and forms monomer if heated to $400^{ \circ  }C$.
(iii) Its aqueous solution turns blue litmus to red and gives a white precipitate with $AgNO _3$ solution, which is soluble in $NH _4OH$.
Addition of $NH _4OH$ and NaOH separately to the solution of $A$ gives a gelatinous precipitate which in however soluble in excess of $NaOH$.
The compound $A$ is:

  1. $Al(OH) _3$

  2. $Al _2Cl _6$

  3. $Al _2O _3$

  4. $Al _2(SO _4) _3$


Correct Option: B
Explanation:

(B) Compound $B$ is ${Al} _{2}{Cl} _{6}$.

(i) ${Al} _{2}{Cl} _{6}$ is a white solid, exists as dimer and fumes in wet air.
(ii) It sublimes at 180°C and converts into $Al{Cl} _{3}$ at 400°C.
(iii) Its aqueous solution is acidic, so it turns blue litmus to red.
${Al} _{2}{Cl} _{6} + 6{H} _{2}O \; \longrightarrow \; \underset{\text{(weak base)}}{2Al{(OH)} _{3}} + \underset{\text{(strong acid)}}{6HCl}$.
It gives white ppt. with $AgN{O} _{3}$ which is soluble in $N{H} _{4}OH$.
(iv)$Al{Cl} _{3} + 3NaOH \; \longrightarrow \; \underset{\text{(white ppt.)}}{2Al{(OH)} _{3}} + 3NaCl$
$Al{(OH)} _{3} + \underset{\text{(excess)}}{NaOH} \; \longrightarrow \; \underset{\text{(soluble)}}{NaAl{O} _{2}} + 2{H} _{2}O$
But in case of $N{H} _{4}OH$, the ppt. of $Al{(OH)} _{3}$ does not dissolve in excess of $N{H} _{4}OH$.

Substance which readily undergo hydrolysis are :

  1. $AlCl _3$

  2. $BCl _3$

  3. $SiCl _4$

  4. $PbCl _4$


Correct Option: A,B,C,D
Explanation:
All the given compounds readily undergo hydrolysis.
$B{ Cl } _{ 3 }+3{ H } _{ 2 }O\longrightarrow B{ \left( OH \right)  } _{ 3 }+3HCl$
$Al{ Cl } _{ 3 }+3{ H } _{ 2 }O\longrightarrow Al{ \left( OH \right)  } _{ 3 }+3HCl$
$Si{ Cl } _{ 4 }+2{ H } _{ 2 }O\longrightarrow Si{ O } _{ 2 }+4HCl$
$Pb{ Cl } _{ 4 }+2{ H } _{ 2 }O\longrightarrow Pb{ O } _{ 2 }+4HCl$
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