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# The mass of P4O10 that will be obtained from the reaction of 1.33 gram of P4 and 5.07 of oxygen is?

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• The mass of P4O10 that will be obtained from the reaction of 1.33 gram of P4 and 5.07 of oxygen is?  ## Answers

Answer #1 | 10/08 2015 21:24 Supposing the missing units on "5.07" to be "grams": (1.33 g P4) / (123.895048 g P4/mol) = 0.010735 mol P4 (5.07 g O2) / (31.99886 g O2/mol) = 0.15844 mol O2 P4 + 5 O2 → P4O10 0.010735 mole of P4 would react completely with 0.010735 x (5/1) = 0.053675 mole of O2, but there is more O2 present than that, so O2 is in excess and P4 is the limiting reactant. (0.010735 mol P4) x (1 mol P4O10 / 1 mol P4) x (283.8890 g P4O10/mol) = 3.05 g P4O10
Positive: 69.230769230769 %
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Positive: 50 %
Answer #3 | 10/08 2015 14:24 Supposing the missing units on "5.07" to be "grams": (1.33 g P4) / (123.895048 g P4/mol) = 0.010735 mol P4 (5.07 g O2) / (31.99886 g O2/mol) = 0.15844 mol O2 P4 + 5 O2 → P4O10 0.010735 mole of P4 would react completely with 0.010735 x (5/1) = 0.053675 mole of O2, but there is more O2 present than that, so O2 is in excess and P4 is the limiting reactant. (0.010735 mol P4) x (1 mol P4O10 / 1 mol P4) x (283.8890 g P4O10/mol) = 3.05 g P4O10
Positive: 50 %

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