What is the balanced chemical equation when barium chloride react with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate?

Last updated at June 30, 2021 by

Answer

(ii) Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride

The skeletal chemical equation of the above reaction is

BaCl 2 + Al 2 (SO 4 ) 3   →  BaSO 4 + AlCl 3

1. Number of atoms of Oxygen in reactants ≠ number of atoms of Oxygen in products
   To balance number of Oxygen,
   We multiply the BaSO 4 in product by 3 (Oxygen is present in BaSO4) The chemical equation becomes

   BaCl 2 + Al 2 (SO 4 ) 3   →  3 BaSO 4 + AlCl 3

   
   Now, 

1. Now,
   Number of atoms of Sulphur in reactants = number of atoms of Sulphur in products
   Number of atoms of Oxygen in reactants = number of atoms of Oxygen in products
   But, Number of other atoms in reactants ≠ number of other atoms in products To balance, number of Aluminium,
   We multiply AlCl 3 in product by 2 (as it contains Al)
   The chemical equation becomes

BaCl 2 + Al 2 (SO 4 ) 3   →  3 BaSO 4 + 2 AlCl 3

Now,

1. Now,
   Number of atoms of Barium in reactants ≠ number of atoms of Barium in products
   Number of atoms of Chlorine in reactants ≠ number of atoms of Chlorine in products
   To balance,
   We multiply BaCl 2 in reactant by 3
   The chemical equation becomes
   3 BaCl 2 + Al 2 (SO 4 ) 3   →  3 BaSO 4 + 2 AlCl 3

Now, all the elements are balanced and the balanced equation is

3 BaCl 2 + Al 2 (SO 4 ) 3   →  3 BaSO 4 + 2 AlCl 3

3 BaCl 2 + Al 2 (SO 4 ) 3   →  3 BaSO 4 + 2 AlCl 3

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Answer

Hint: Barium chloride stands for$BaC{{l}_{2}}$, aluminum sulphate stands for $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$, barium sulphate stands for $BaS{{O}_{4}}$ and last aluminum chloride stands for $AlC{{l}_{_{3}}}$. Arrange them, chemical reactions and balance them.The symbolic representation of barium is $Ba$.The symbolic representation of chloride is $Cl$.The symbolic representation of aluminum is $Al$.The symbolic representation of sulphate is $S{{O}_{4}}$. Complete step-by-step answer:Barium chloride is represented as $BaC{{l}_{2}}$ barium has oxidation state and chlorine has oxidation state. Therefore 1 barium combines with 2 chlorides.Aluminum sulphate is represented by $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$ as aluminum has oxidation state and sulphate has oxidation state. So, 3 aluminum reacts with 3 molecules of sulphate.When barium chloride reacts with aluminum sulphate it gives aluminum chloride and barium sulphate as a product.Barium sulphate is represented as $BaS{{O}_{4}}$ barium has oxidation state and sulphate has oxidation state. Therefore 1 barium combines with 1 Sulphur. Aluminium chloride is represented by $AlC{{l}_{_{3}}}$ as aluminum has oxidation state and chlorine has oxidation state. So, 1 aluminum reacts with 3 chlorines.So overall reaction is:$BaC{{l}_{2}}+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}\to AlC{{l}_{3}}+BaS{{O}_{4}}$Now, balancing the equation as the number of elements as they are not equal on both sides.So the balanced equation is: $3BaC{{l}_{2}}+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}\to 2AlC{{l}_{3}}+3BaS{{O}_{4}}$Now labeling the physical state of each reactant and product:$3BaC{{l}_{2}}(aq)+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}(aq)\to 2AlC{{l}_{3}}(aq)+3BaS{{O}_{4}}(s)$It is given that barium sulphate precipitates therefore it is solid and others are aqueous in their physical state. Note: Chemical equation is written by the chemical name of each compound present as reactant or product arranging them in an equation followed by balancing of equation as stoichiometry should be the same. After balancing the equation, the physical state of each reactant and product is written.