# How do you balance PbSO_4 → PbSO_3 + O_2?

Jun 2, 2016

2 PbSO4→2 PbSO3 + O2

#### Explanation:

With problems like this, I would do an atom inventory to see how many atoms of each element are present on both sides of the reaction arrow.

Initially, you have 1 atom of Pb on both the reactants and products side, followed by 1 atom of sulfur on both sides, and 4 atoms of oxygen on the reactants side and 5 atoms of oxygen on the products side.

Place a coefficient of two in front of the PbSO4 to obtain 2 atoms of Pb and S, and 8 atoms of O.

I started with 2 because I based it off the number of oxygen atoms on both sides of the reaction arrow. I knew that if there's a 2 in front of PbSO₄ , there will be 8 oxygen atoms on the left hand side. And if you place 2 on the right hand side in front of $P b S {O}_{3}$, you'll end up with 6 oxygen atoms, plus the two that you already have from
${O}_{2}$, which happens to balance all the atoms.

I basically focused on the oxygen atoms since lead and sulfur has a coefficient of 1. Next to balance the Pb and S on the products side, place a coefficient of two in front of $P b S {O}_{3}$ to obtain 2 atoms of Pb and S, and 6 atoms of O. The last two oxygen atoms come from the ${O}_{2}$, which gives you a total of 8 oxygen atoms.

Now it's balanced! Hope I helped.