# How can I balance this equation? ___ AlBr3 + ____ K2SO4 ---> ____ KBr + ____ Al2(SO4)3

May 30, 2014

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

How I balance equations is by first of all dividing the equation into two parts - reactants and products.

Reactants
Al = 3
Br = 3
K = 2
S = 1
O = 4

Products
Al = 2
Br = 1
K = 1
S = 3
O = 12

Remember all atoms within the brackets get multiplied by the number outwith them.

Step 1
Put a 6 before KBr

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

Step 2
Put a 2 before $A l B {r}_{3}$
- this has now balanced the number of Aluminium and Bromine atoms.

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

Step 3
Put a 3 before ${K}_{2} S {O}_{4}$

This balances the Potassium, Sulphur and Oxyen

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

Now check the reactants and products:

Reactants
Al - 2
Br - 6
K - 6
S - 3
O - 12

Products
Al - 2
Br - 6
K - 6
S - 3
O - 12

We have now successfully balanced the equation

I hope this helps!