3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s)
The unbalanced equation is:
ZnBr2 (aq) + Al (s) ==> AlBr3 (aq) +Zn (s)
First, count the atoms of each element on each side of the equation:
Zn 1,1
Br 2,3
Al 1,1
The zinc and aluminum, but the bromine doesn't match with 2 and 3. So look for the least common multiple of 2 and 3 which is 6 and adjust the quantities on both sides to have 6 bromine atoms on both sides. Do this by having 3 zinc bromide on the left and 2 aluminum bromide on the right, getting:
3 ZnBr2 (aq) + Al (s) ==> 2 AlBr3 (aq) +Zn (s)
Now check the atom counts again for both sides:
Zn 3,1
Br 6,6
Al 1,2
Now bromine matches, but zinc and aluminum doesn't. But it's easy enough to add an extra aluminum to the left and 2 more zinc to the right. Giving:
3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s)
Now check the atom counts again:
Zn 3,3
Br 6,6
Al 2,2
And they match. So the balanced equation is:
3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s)
