Question

Tetraphosphorus hexaoxide (MM = 219.9 g/mol) is formed by the reaction of phosphorus with oxygen gas. P4(s) + 3O2(g) → P4O6(s) If 42.37 g of oxygen reacts with excess phosphorus and 65.5 g of P4O6 are produced, what is the percent yield for the reaction?

Answer 1

mass O₂ consumed = 42.37 g

moles O₂ consumed = (mass O₂ consumed) / (molar mass O₂)

moles O₂ consumed = (42.37 g) / (32.0 g/mol)

moles O₂ consumed = 1.324 mol

moles P₄O₆ formed = (moles O₂ consumed) * (1 moles P₄O₆ / 3 moles O₂)

moles P₄O₆ formed = (1.324 mol) * (1 / 3)

moles P₄O₆ formed = 0.441 mol

Theoretical yield P₄O₆ = (moles P₄O₆ formed) * (molar mass P₄O₆)

Theoretical yield P₄O₆ = (0.441 mol) * (219.89 g/mol)

Theoretical yield P₄O₆ = 96.97 g

Percent yield = (Actual yield / Theoretical yield P₄O₆) * 100

Percent yield = (65.5 g / 96.97 g) * 100

Percent yield = (0.6686) * 100

Percent yield = 66.9 %