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2Na3PO4 + 3CaCl2 ----> Ca3 (PO4)2 + 6NaCl

how many moles of CaCl2 remain if .10 mol Na3PO4 and .40 mol CaCl2 are used?


When amounts of both materials are listed one must worry about which is the limiting reagentalthough the problem seems to suggest that Na3PO4 is the limiting reagent. We can take them in order. I don't have great eye sight but I think your numbers are 0.10 mol Na3PO4 and 0.40 mol CaCl2.

0.10 mol Na3PO4 x [3 mol CaCl2/2 mol Na3PO4] =0.15 mol CaCl2 and that subtracted from 0.40 = 0.25 mol CaCl2 remaining. You can check this out; all of the Na3PO4 should have been used and it is the limiting reagent.

0.40 mol CaCl2 x [2 mol Na3PO4/3 mol CaCl2 = 0.267 mol Na3PO4. We didn't have that much initially; therefore, all of it must have been consumed. I hope this helps.

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Hawk

  1. 0.25

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    bobbi

  2. above me is wrong

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    bobbi

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