Chemistry help please

Question

How many grams of oxygen are required to react with 38.0 g of lead(II) sulfide?

How many grams of sulfur dioxide can be produced when 66.0 g of lead(II) sulfide reacts?

How many grams of lead(II) are used to produce 124 g of lead(II) oxide?

Answer 1

The problem is a little vague about the reactions. For part 1 I'll assume it's PbS + O₂ --> Pb + SO₂, then (38.0g PbS)*(1 mol PbS / 239.3g PbS)*(1 mol O₂ /1 mol PbS)*(32.0g O₂ /1 mol O₂) = 5.08g O₂.

Part 2, with the same reaction: (66.0g PbS)*(1 mol PbS / 239.3g PbS)*(1 mol SO₂ /1 mol PbS)*(64.1g SO₂ /1 mol SO₂) = 17.7g SO₂.

Part 3, I assume 2Pb + O₂ --> 2PbO, so (124g PbO)*(1 mol PbO /223.2g PbO)*(2 mol Pb /2 mol PbO)*(207.2g Pb /1 mol Pb) =115g Pb.