Question

1. In order to synthesize diphosphorus pentoxide, 12.0 g of P react with 35.0 g of O2(g), resulting in the production of 21.2 g of P205. Write the balanced chemical equation, then determine the limiting reactant, the theoretical yield, and the percent yield.

Answer 1

Solution 1-> The balanced equation betrocen P and Oy for the formation of Pros is 4P + 502 2P₂Os 12.0 Mais of P used Molar mais of Р 30.97 glmot. Mars 12.0 PS Moles of = 0.387 moles Molar mans 30.97 Mars used - 35.0g Molar mous = 32 glmol. 35.0 32 1.094 moles Molar mous off of O2 of O₂ Moles of O₂ Mars of O2 6702 From the above chemical equation, it 4 moles of P reacts with ₂ i. I mole of P reacts with 02 :: 0.387 moles of P react with clear that 5 moles -5 moles. 4 0, & X0:387 5 4 = 0.484 moles But we have 1.094 moles of O2, which is greater than 0.484 moles. This means O₂ is present large quantity compared to P. So, is the limitting reagent. Now, from the balanced chemical equation above 4 moles of P reaat wou gives P2O5 2 moles.

2 moles. 2 x0'384 4 .. 1 mole of P gives P2O5 = 2 4 0.387 moles of P gives P2O5 = Moles of P2Os Now, Molar mars of P2Os = 141.94 glonot i. Theoretical Yield of P205 = Moles of P2 OsX Molar mans of P2Os = 0.192 moles -0.192 X 141.94 - 27.25 77.259 Experiment field of P2O5 = 21.2g Percent Yield Experimental Yield theoretical Yield x100 21.2 27.25 x 100 = 77.79%