5.04 g of lead combines directly with 0.78 g of oxygen to form lead dioxide (PbO2). Lead dioxide is also produced by heating lead nitrate and it was found that the percentage of oxygen present in lead dioxide is ________ . Use these data to illustrate the law of constant composition.

Step 1: To calculate the percentage of oxygen in firstexperiment.Weight of peroxide formed= 5.04 + 0.78 - 5.82 g.5.82 g of lead dioxide contains 0.78 g of oxygen.'. 100 g of lead dioxide will contain oxygen0.785.82×100=13.41That is, oxygen present : 13.41%Step 2: To compare the percentage of oxygen in both theexperiments.Percentage of oxygen in PbO2 in the first experiment= 13.41Percentage of oxygen in PbO2 in- the second experimentSince the percentage composition =13.41 @of oxygen in both thesamples of PbO2 is the same, the above data illustrate the lawof constant composition.