First you identify the excess reagent. I do that the long way by taking the reactants, one at a time, to calculate the amount of product formed.
mols Na3PO4 = M x L = about 0.131
mols CuCl2 = M x L = about 0.000969,
How much NaCl could be formed with Na3PO4 and all of the CuCl2 needed. That is 0.131 mols Na3PO4 x (6 mols NaCl/2 mols Na3PO4) = 0.131 x (6/2) = approx 0.0394.
How much NaCl could be formed with CuCl2 and all of the Na3PO4 needed. That is 0.00969 x (6 mols NaCl/3 mols CuCl2) = 0.00969 x (6/3) = 0.0194
In limiting reagent problems the smaller number wins so CuCl2 is the limiting reagent.
Now follow the same procedure in determining mols Na3PO4 used by the 0.00969 mols CuCl2 and subtract from the initial amount to find the amount not used.
Consider the following reaction:
2Na3PO4(aq)+3CuCl2(aq) = Cu3(PO4)2(s)+ 6NaCl(aq)
If 75mL of .175 M Na3PO4 combined with 95mL of .102M CuCl2, determine the concentration of the excess reagent.
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