For the given reaction, what volume of O2 would be required to react with 4.8 L of PCl3 , measured at the same temperature and pressure? 2PCl3(g)+O2(g)⟶2POCl3(g)

Answer 1

When the temperature and pressure are constant, you can use volume (L) the same way you would use moles in a stoichiometry problem. Thus...

2PCl₃(g) + O₂(g) ⟶ 2POCl₃(g) ... balanced equation

4.8 L PCl₃ x 1 L O₂ / 2 L PCl₃ = 2.4 L O₂ needed