I work these problems the long way.
First write and balance the equation.
3CaCl2 + 2Na3PO4 ==> Ca3(PO4)2 + 6NaCl
Convert 18.5g CaCl2 to moles. moles = grams/molar mass. Using the coefficients in the balanced equation, convert mols CaCl2 to moles Ca3(PO4)2. The answer you obtain is the mols Ca3(PO4)2 produced by 18.5 g CaCl2 if you had an excess of the Na3PO4.
Next, do the same process for Na3PO4. This will give you the mols Ca3(PO4)2 produced if you had 25.7 g Na3PO4 and an excess of CaCl2.
The answers for mols Ca3(PO4)2 produced by the two scenarios will not be the same which means one of them is not right; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
The problem doesn't ask for it but if you want grams produced, then mols Ca3(PO4)2 x molar mass = grams.
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