# Question #7132d

Jan 21, 2016

$2 A l B {r}_{3} + 3 {K}_{2} S {O}_{4} \to 6 K B r + A {l}_{2} {\left(S {O}_{4}\right)}_{3}$

#### Explanation:

Let $a , b , c \mathmr{and} d$ be the number of molecules in the balanced equation as shown below.

$a A l B {r}_{3} + b {K}_{2} S {O}_{4} \to c K B r + \mathrm{dA} {l}_{2} {\left(S {O}_{4}\right)}_{3}$

Inspection reveals that the on the right hand side of the equation number of radical $\left(S {O}_{4}\right)$ is the highest, i.e., $3$. Hence take $d$ as a lowest integer.
Therefore, with $d = 1$, the equation becomes

$a A l B {r}_{3} + b {K}_{2} S {O}_{4} \to c K B r + A {l}_{2} {\left(S {O}_{4}\right)}_{3}$
To balance 3 radicals $S {O}_{4}$, on the left hand side $b = 3$.
With this the equation becomes

$a A l B {r}_{3} + 3 {K}_{2} S {O}_{4} \to c K B r + A {l}_{2} {\left(S {O}_{4}\right)}_{3}$
Now balancing $6$ atoms of $K$ on the left hand side, $c = 6$
we get

$a A l B {r}_{3} + 3 {K}_{2} S {O}_{4} \to 6 K B r + A {l}_{2} {\left(S {O}_{4}\right)}_{3}$
To find the value of $a$, you may either balance $A l \mathmr{and} B r$. We obtain $a = 2$ and get the balanced equation as in the Answer.