Phosphine gas oxidizes spontaneously on exposure to air, as represented by: 2 PH₃(g) + 3 O₂(g) → P₂O₃(s) + 3 H₂O(g)
What volume of oxygen gas, at 720 torr and 37°C, would be consumed in the formation of 22.0 g of P₂O₃ by this process?

Question

Quizplus · Accepted Answer

The balanced chemical equation shows that 2 moles of PH3 react with 3 moles of O2. Therefore, the mole ratio of PH3 to O2 is 2:3. The molar mass of P2O3 is 142 g/mol. Therefore, 22.0 g of P2O3 is equal to 0.155 moles of P2O3. Using the mole ratio, we can determine that 0.155 moles of P2O3 is produced from 0.233 moles of O2. The ideal gas law can be used to calculate the volume of O2 at 720 torr and 37°C:PV = nRTV = nRT/PV = (0.233 mol)(0.0821 Latm/(molK))(310 K)/(720 torr)V = 16.1 L

Phosphine Gas Oxidizes Spontaneously on Exposure to Air, as Represented