Final answer:
The percent yield of the reaction when 2.80 g of P reacts with excess oxygen to produce 4.89 g of P2O5 is approximately 76.24%, which is not listed in the given options.
Step-by-step explanation:
To determine the percent yield of the reaction where 2.80 g of P reacts with excess oxygen to produce 4.89 g of P2O5, we first need to calculate the theoretical yield and then use it to find the percent yield. The balanced chemical equation given is: 4P + 5O2 → 2P2O5. The molar mass of phosphorus (P) is about 30.97 g/mol and that of diphosphorus pentoxide (P2O5) is 141.94 g/mol.
First, we convert grams of P to moles:
(2.80 g P)/(30.97 g/mol) ≈ 0.0904 mol P.
From the balanced equation, 4 moles of P yield 2 moles of P2O5, so 0.0904 mol P would yield:
(0.0904 mol P) * (2 mol P2O5 / 4 mol P) ≈ 0.0452 mol P2O5.
Now we convert moles of P2O5 to grams to find the theoretical yield:
0.0452 mol P2O5 * (141.94 g/mol) ≈ 6.416 g P2O5.
Finally, the percent yield is calculated as follows:
Percent Yield = (Actual Yield / Theoretical Yield) * 100%
Percent Yield = (4.89 g / 6.416 g) * 100% ≈ 76.24%%.
This is not one of the options provided, which suggests there might be a mistake in the calculation or in the options given. If the student had options matching this result, they should choose the one closest to 76.24%.