Please help I’m so confused

Question

When lead (II) sulfide is burned in air, lead (II) oxide and sulfur dioxide are produced. If 0.890 moles of sulfur dioxide were produced, how many moles of oxygen gas were required to react with the lead (II) sulfide?

Answer 1

Lead(II) sulfide = PbS

Oxygen = O₂

Lead(II) oxide = PbO

Sulfur dioxide = SO₂

Write the balanced equation for the reaction taking place:

2 PbS + 3 O₂ ==> 2 PbO + 2 SO₂ ... balanced equation

Now simply use the stoichiometry (mole ratios) of the balanced equation and dimensional analysis:

0.890 mols SO₂ x 3 mol O₂ / 2 mol SO₂ = 1.34 mols O₂ needed (3 sig. figs.)