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

First, calculate the moles of P4O10 formed from 22.0 g. The molar mass of P4O10 is 283.89 g/mol. Moles of P4O10 = 22.0 g / 283.89 g/mol = 0.0775 mol. From the balanced equation, 3 moles of O2 are consumed to produce 1 mole of P4O10, so 0.0775 mol of P4O10 requires 0.2325 mol of O2. Using the ideal gas law PV = nRT and converting the temperature to Kelvin (37°C + 273 = 310 K) and pressure to atm (720 Torr / 760 = 0.9474 atm), the volume of O2 can be calculated as V = nRT/P = (0.2325 mol)(0.0821 L·atm/mol·K)(310 K) / (0.9474 atm) ≈ 16.1 L.

Phosphine Gas Oxidizes Spontaneously on Exposure to Air,as Represented By