Question

Tetraphosphorus hexoxide is formed by the reaction of phosphorus with oxygen gas. P4(s) 302(g)P406(s If a mixture of 75.3 g of phosphorus and 38.7 g of oxygen produce 43.3 g of PO%, what is the rcent yield for the reaction? A. 32.4% B. 38.0% C. 48.8% D. 16.3% E. 57.5%

Please slove step by step

Answer 1

Option (C) 48.8% P_4 (s) + 3 O_2 (g) rightarrow P_4 O_6 (s) P_4 O_6 molecular weight = 219.9 g/mol step I: find the Number of moles of P_4, O_2, P_4 O_6 P_4 moles = given P_4 weight /Molecular weight of P_4 = 75.3/123.90 = 0.60774 moles. O_2 moles = given O_2 weight /Molecular weight of O_2 = 38.79 /32 = 1.209 moles P_4 O_6 moles = given P_4 O_6 weight/Molecular weight of P_4 O_6 = 43.3/219.9 = 0.196 moles \r\n step II: calculate the Theoretical yield (100% yield) The limiting is oxygen. 3 moles of oxygen produces 1 mole of P_4 O_6 1.209 moles of oxygen produces moles of P_4 O_6 = 1.204/3 = 0.403 moles of P_4 O_6 = 0.403 times 219.9 gms of P_4 O_6 = 88.619 g ms of P_4 O_6 Theoretical yield = 88.619 g m s (100% yield) step III: % of yield = practical weight/theoretical weight times 100