Martin S. answered • 04/15/20
Patient, Relaxed PhD Molecular Biologist for Science and Math Tutoring
The reaction is Na3PO4 + CuCl2 → Cu3(PO4)2 + NaCl.
To determine which reactant is limiting, determine the relative molar quantity of each reactant, and compare that to how much would be needed in the reaction. The reactant that would be completely used up is the limiting reactant.
First, add the atomic masses of the elements Na3PO4 to find the molecular mass, or the grams per mole. The atomic mass of Na is 23, for P it is 31, for O it is 16. Multiply each of those by the number of atoms per molecule of Na3PO4 and you get (3x23) + 31 + (4x16) = 164 g/mol. You started with 1.785 g Na3PO4 , so dividing that by the molecular mass and you had 1.785 g/164 g/mol = 1.09 x 10-2 moles of Na3PO4. Each molecule of Na3PO4 has one phosphate, so that is also the number of moles of phosphate you started with.
That was added to 20 mL of a 0.678 mol/L solution of CuCl2. The amount of Cu that you started with is found by multiplying the molar concentration times the volume. 20 mL is 2 x 10-2 L, so multiplying that by 0.678 mol/L gives you (2 x 10-2 L) x (0.678 mol/L) = 1.354 x 10-2 moles CuCl2, which is also the number of moiles of Cu you started with since there is one atom of Cu per molecule of CuCl2,
In the product, Cu3(PO4)2, there are 3 moles of Cu for every 2 moles of PO4, and that ratio determines how much of each reactant could be used. You need 1 1/2 times as many moles of Cu as moles of PO4 in order to use the PO4 to completion ( a 3:2 ratio).. So to use all of the PO4, you would have needed 3/2 x (1.09 x 10-2) moles of Cu, or 1.64 x 10-2 moles. But you only had 1.354 x 10-2 moles CuCl2 to start with in the solution, which was not enough to use all the PO4. So CuCl2 was the limiting reactant.
Hope that helps
