Question

$BariumChloride+AluminiumSulphate\to BariumSulphate+AluminiumChloride$

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

• A. $3BaC{l}_3$(aq) + $Al\left(S{O}_4\right)\left(s\right)$ $\to$ $B{a}_2\left(S{O}_4{)}_3\right(s)$ + $2AlC{l}_2$
• B. $2BaC{l}_3$(aq) + $3Al\left(S{O}_4\right)\left(s\right)$ $\to$ $B{a}_2\left(S{O}_4{)}_3\right(s)$ + $3AlC{l}_2$
• C. $3BaC{l}_2$(aq) + $A{l}_2\left(S{O}_4{)}_3\right(s)$ $\to$ $2Ba\left(S{O}_4\right)\left(s\right)$ + $3AlC{l}_3$
• D. $3BaC{l}_2$ + $A{l}_2(S{O}_4{)}_3$ $\to$ $3BaS{O}_4$ + $2AlC{l}_3$

Answer 1

The correct option is D

$3BaC{l}_2$ + $A{l}_2(S{O}_4{)}_3$ $\to$ $3BaS{O}_4$ + $2AlC{l}_3$

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

The skeleton reaction is $BaC{l}_2$ + $A{l}_2(S{O}_4{)}_3$ $\to$ $BaS{O}_4+AlC{l}_3$

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,$BaC{l}_2$ + $A{l}_2(S{O}_4{)}_3$ $\to$ $BaS{O}_4+2AlC{l}_3$

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.

$BaC{l}_2$ + $A{l}_2(S{O}_4{)}_3$ $\to$ $3BaS{O}_4+AlC{l}_3$

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.

$3BaC{l}_2$ + $A{l}_2(S{O}_4{)}_3$ $\to$ $3BaS{O}_4+2AlC{l}_3$

The balanced reaction is,

$3BaC{l}_2$ + $A{l}_2(S{O}_4{)}_3$ $\to$ $3BaS{O}_4+2AlC{l}_3$

$BaC{l}_2$ and $AlC{l}_3$ are soluble in water so written as (aq). $BaS{O}_4$and $A{l}_2(S{O}_4{)}_3$ are insoluble in water so written as (s).