$B a r i u m C h l o r i d e + A l u m i n i u m S u l p h a t e \to B a r i u m S u l p h a t e + A l u m i n i u m C h l o r i d e$

Which of the following options correctly represents the balanced chemical equation for the above reaction?

$3 B a C l_{2}$ + $A l_{2} (S O_{4})_{3}$ $\to$ $3 B a S O_{4}$ + $2 A l C l_{3}$

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$3 B a C l_{3}$ (aq) + $A l (S O_{4}) (s)$ $\to$ $B a_{2} (S O_{4})_{3} (s)$ + $2 A l C l_{2}$

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$2 B a C l_{3}$ (aq) + $3 A l (S O_{4}) (s)$ $\to$ $B a_{2} (S O_{4})_{3} (s)$ + $3 A l C l_{2}$

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$3 B a C l_{2}$ (aq) + $A l_{2} (S O_{4})_{3} (s)$ $\to$ $2 B a (S O_{4}) (s)$ + $3 A l C l_{3}$

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Solution

The correct option is A
$3 B a C l_{2}$ + $A l_{2} (S O_{4})_{3}$ $\to$ $3 B a S O_{4}$ + $2 A l C l_{3}$

Barium Chloride + Aluminium Sulphate→ Barium Sulphate +Aluminium Chloride.

The skeleton reaction is $B a C l_{2}$ + $A l_{2} (S O_{4})_{3}$ $\to$ $B a S O_{4} + A l C l_{3}$

Consider Aluminium Sulphate compound first because it has more number of atoms, 2 atoms of Al are present on reactant side and 1 atom on product side, to balance Al, take the coefficient of AlCl₃ on product side as 2.

So the equation becomes, $B a C l_{2}$ + $A l_{2} (S O_{4})_{3}$ $\to$ $B a S O_{4} + 2 A l C l_{3}$

Now consider the set (SO₄) instead of S and O separately because it is present in reactant and product side in the same form. There are 3 sets on reactant side and 1 set on product side, to balance this, take the coefficient of BaSO₄ as 3.

$B a C l_{2}$ + $A l_{2} (S O_{4})_{3}$ $\to$ $3 B a S O_{4} + A l C l_{3}$

Now consider Ba, there are 3 atoms on product side and 1 atom on reactant side, to balance it, take the coefficient of BaCl₂ as 3. The Cl atoms are also balanced.

$3 B a C l_{2}$ + $A l_{2} (S O_{4})_{3}$ $\to$ $3 B a S O_{4} + 2 A l C l_{3}$

The balanced reaction is,

$3 B a C l_{2}$ + $A l_{2} (S O_{4})_{3}$ $\to$ $3 B a S O_{4} + 2 A l C l_{3}$

$B a C l_{2}$ and $A l C l_{3}$ are soluble in water so written as (aq). $B a S O_{4}$ and $A l_{2} (S O_{4})_{3}$ are insoluble in water so written as (s).

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