Question

Consider a situation in which 161 g of P4 are exposed to 176 g of O2.

What is the percent yield if the actual yield from this reaction is 259 g ? Express your answer to three significant figures and include the appropriate units.

Answer 1

Answer 1.61 g of Py are exposed to 176g of O2 Reaction- Put O2 → Role PGO 10 Balance equation Py + 502 = el PqO10 Molecular weight of Py = 123.9 g/mol moles of Py = 16 lg 12310 gand 21.3 mol molecular weight of O₂ = 32 g/mol Moles of a q a 176 g. 32 g/mol -515 mol 515 mol O2 4.23 mol O2/mol Py 1.3 mol Py actual 14 23 moldz: 1 mol Py from balance equation we get Ideal 5 mol O₂: 1 mol py actual <ideal So, Py is limiting reactant

Put 502 - Py010 1.3 mol Py 113 mol PGO 10 molecular weight of P40 10 = 283.9 g/ ono 103 mol PuO10 = 1'3x2839 g 369107 g Actual yield of the reaction = 259 g Theoritical yield of the reaction - 369007g % yield a actual yield theoritical yield x 100% , 259 x 100 369.07 * 100% - 70.18%