Question

Review / Constants Periodic Table Part C In Part A, you found the number of moles of product 2.60 mol P2Os) formed from the given amount of phosphorus and excess oxygen. In Part B, you found the number of moles of product (2.20 mol P2Os ) formed from the given amount of oxygen and excess phosphorus. Now, determine the number of moles of P20 is produced from the given amounts of phosphorus and oxygen. Express your answer to three significantigures and include the appropriate units. View Available Hint(s) Value Units lo Submit Previous Answers Cal X Incorrect; Try Again; 2 attempts remaining first ants cying You were asked to find the number of moles of P.Os.

Answer 1

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+ Buclig) Answer: Bra (9) + Och (9) Broclo when oxygen is in excess Py +502 2 P₂ os for production of a moles of P2O5 - 1 mole Py is used for production of 2.6 moles of P2O5 2. 6x1 mole Py is used Mpy = 113 mol mass of Py = No. of moles & Molare mass of Pu =113 x 31 x 4 g - 176.89 3 mol is in excess when Phosphorus Py +502 - 2 P204 for 2 mole q P204 formation - smoles q On is used for 2.2 mole of Prou formation - $x2.2 moles of Oz is used no = 5.5 mol Now, Py + 502 - 2 P205 for limiting reagent which has less value calculate mole a s stoichiometry act of limitin reagent.

for Py = 13 = 13 for O2 = 5.5 =10 Hence, O2 is limiting reagent. Reaction as por o2 proceed for 5 mole of oz for 5.5 mole of oz 2 moles of P2O5 is formed 2x5.5 moles of P2 Os is formed, por = 2. 20 mol.

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