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judd judd
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13 years ago
What is the limiting reactant for the following reaction given we have 3.4 moles of Ca(NO3)2 and 2.4 moles of?
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#1
13 years ago
3Ca(NO3)2 + 2Li3PO4 ----> 6LiNO3 + Ca3(PO4)2

3 mol Ca(NO3)2 reacts with 2 mol Li3PO4

Assume all the Ca(NO3)2 is used up:
3.4 mol Ca(NO3)2 x (2 mol Li3PO4 / 3 mol Ca(NO3)2 ) = 2.27 mol Li3PO4

Since there are more moles of Li3PO4 than the 2.27 moles required to use up all the Ca(NO3)2, Li3PO4 is present in excess. This means the Ca(NO3)2 is the limiting reagent.

You could also assume that all the Li3PO4 is used up.
2.4 mol Li3PO4 x (3 mol Ca(NO3)2 / 2 mol Li3PO4) = 3.6 mol Ca(NO3)2.

It would take 3.6 moles of Ca(NO3)2 to use up all the Li3PO4, but there are only 3.4 moles of Ca(NO3)2 present. Thus, Ca(NO3)2 must be the limiting reagent.
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