Okay, so here's the approach to this one -
You know that you started with 4.5 moles of PbCl2. That give you how many moles of Pb atoms going into the reaction?
At one Pb atom per molecule of PbCl2, and 4.5 moles of PbCl2, that gives you.... 4.5 moles of Pb.
Looking at the right side of the equation, you see that you come out with one molecule of PbO for each atom of Pb.
At one Pb atom per molecule of PbO, and 4.5 moles of Pb, that gives you... 4.5 moles of PbO.
How much does each mole of PbO weigh?
Add up the atomic weight of Pb with the atomic weight of O, and it's that many grams.
If I recall correctly, Oxygen is close to 16 and Lead is around 205 to 207... use the numbers off the periodic chart and round to 1 digit. Let's say 16 + 206.8 = 222.8 to finish the example.
Multiply 222.8 grams per mole by 4.5 moles to get 1002.6 grams of PbO.
Just to show you a variation, let's suppose that the reaction had been creating red lead oxide (Pb3O4) instead of black (PbO)
3PbCl2 + 4O2 ---> Pb3O4 + 3Cl2
We still have 4.5 moles of PbCl2 molecules going in. So we have 4.5 moles of PB atoms coming out.
However, it takes 3 Pb atoms to make one molecule of Pb3O4. That means we only get 4.5/3 = 1.5 moles of Pb3O4 molecules coming out.
One mole of Pb3O4 has a weight, using my made up numbers of 206.8 and 16, 3*206.8 + 4*16 = 684.4 grams per mole. We have 1.5 moles, so the resulting oxide has a total weight of 1026.6 grams.
Since they had the same number of lead atoms in the precipitates, the difference between the two weights is the difference in the amount of bound up oxygen. By inspection of the formulas, you can see that there is one extra oxygen atoms per three lead atoms, and you know there were 4.5 moles of lead atoms, so the difference is 1.5 moles of Oxygen atoms, which should be 16*1.5 or 24 extra grams.
1026.6 - 1002.6 = 24, check.
Obviously, you'll need to reference the real atomic weights when you do this kind of problem, but otherwise, that's the method.