What volume of oxygen gas is required to react completely with 2.07 mol of hydrogen sulfide according to the following reaction at 0°C and 1 atm?

Question

hydrogen sulfide (g) + oxygen(g)water (l) + sulfur dioxide(g)

Answer 1

H₂S (g) + O₂ --> H₂O (l) + SO₂

Because this is a 1:1 ratio of Oxygen to Hydrogen Sulfide, you need the same amount of mole of both to fully react both.

we have 2.07 moles of hydrogen sulfide, so we need 2.07 moles of oxygen.

PV = nRT

V=nRT/P

V = (2.07 mol)(0.08 L atm mol^-1K^-1 )(273.15 K)/(1 atm)

V = 45.23 L