Given the equation Pb02 -> Pbo +O2, how many grams of oxygen will be produced if 47.8 g of lead (IV) oxide

(cont.) decompose to form 44.6 g of lead (II) oxide and oxygen gas?

Either yhour figures are wrong or the answer is 47.8 - 44.6 = 3.2g
i.e. it's arithmetic, not chemistry.

PbO2 --> Pb + O2

Moles = Mass / Molar Mass

Mass (PbO2) = 47.8 g

Molar Mass (PbO2) = 239.2 g/mol

Moles (PbO2) = 47.8 / 239.2 = 0.2 mol

Ratio of reaction between PbO2 and O2 is 1:1, so we have 0.2 mol of O2

Mass = Moles x Molar Mass

Moles (O2) = 0.2 mol

Molar Mass (O2) = 32 g/mol

Mass (O2) = 0.2 x 32 = 6.4 g

I've ignored that 47.8 g become 44.6 g, as that is impossible based on conservation laws.