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Equal weights of P and O are heated in a closed vessel producing P203 andP20, in a 1:1 mole ratio. If the limiting component is exhausted, find whichcomponent and what fraction of it is left over?. |
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Answer» Answer: O is exhausted and 3.20% P left over Explanation: 4 P + 3 O2 → 2 P2O3 4 P + 5 O2 → 2 P2O5 Since you are told that the RESULT is equal moles of P2O3 and P2O5, these two equations can be ADDED together to form the overall reaction: 8 P + 8 O2 → 2 P2O3 + 2 P2O5 Then simplifying: 4 P + 4 O2 → P2O3 + P2O5 Take hypothetical samples of 100 grams each of P and O2: (100 g P) / (30.97376 g P/mol) = 3.2285 mol P (100 g O2) / (31.99886 g O2/mol) = 3.1251 mol O2 3.1251 moles of O2 would react completely with 3.1251 x (4/4) = 3.1251 moles of P, but there is more P present than that, so P is in excess and O2 is the LIMITING REACTANT (component), meaning that O2 is exhausted and P is left over. (3.2285 mol P initially) - (3.1251 mol P reacted) = 0.1034 mol P left over (0.1034 mol P) / (3.2285 mol P initially) = 0.0320 = 3.20% P left over |
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