# How would you balance: AlBr3 + K2SO4 --> KBr + Al2(SO4)3?

Nov 27, 2015

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

#### Explanation:

The first thing I notice is that the product side has 3 sulfate groups, while the reactant side has only 1. We can multiply the potassium sulfate group by 3 so there are 3 on either side.

However, this leaves us with 6 potassium atoms on the reactant side and only 1 on the product side, so we multiply the potassium bromide by 6.

Now, we have the imbalance of 6 bromine atoms on the right and only 3 on the right, so we multiply the aluminum bromide on the right by 2. Finally, this leaves 2 aluminum atoms on the reactant side and on the product side, so we are done balancing.