Question

Tetraphosphorus hexaoxide ( ℳ = 219.9 g/mol) is formed by the reaction of phosphorus with oxygen gas. p4(s) + 3O₂(g) → p4o6(s). If a mixture of 75.3 g of phosphorus and 38.7 g of oxygen produce 43.3 g of p4o6, what is the percent yield for the reaction?

Answer 1

The percent yield is calculated using the actual yield of the reaction and the theoretical yield, determined by stoichiometry from the reaction P4(s) + 3O2(g) → P4O6(s). Phosphorus, with an actual yield of 43.3 g of P4O6, is the limiting reactant.

Step-by-step explanation:

The percent yield for the reaction is calculated using the actual yield and the theoretical yield. The actual yield is the amount of product produced in a reaction, while the theoretical yield is the amount of product that could be produced if the reaction were perfect. The formula for percent yield is: Percent Yield = (Actual Yield / Theoretical Yield) × 100%.

To find the theoretical yield, the moles of limiting reactant are determined. Phosphorus here is the limiting reactant since it will produce less product than the provided oxygen would. The moles of phosphorus (P4) and oxygen (O2) are calculated from their given masses and molar masses. Using stoichiometry from the balanced equation P4(s) + 3O2(g) → P4O6(s), the moles of tetraphosphorus hexaoxide (P4O6) that should be produced are found. Then, the theoretical yield in grams is determined by multiplying the moles of P4O6 by its molar mass.

The percent yield is then:

Percent Yield = (43.3 g / Theoretical Yield) × 100%