50.0ml of an aqueous solution of 2.50M H3PO4 are mixed with 75.0 ml of an aqueous solution of 4.00 M Ca(OH)2. The resulting precipitate is washed and isolated using filtration. What mass of Ca3(PO4)2 is obtained?

Answer 1

This is a limiting reagent (LR) problem. I know that becsuer amounts are given for BOTH reactants.
2H3PO4 + 3Ca(OH)2 ==> Ca3(PO4)2 + 6H2O

1. mols H3PO4 = M x L = ?
Do the same for mols Ca(OH)2.

2a. Using the coefficients in the balanced equation, convert mols H3PO4 to mols Ca3(PO4)2.
2b. Do the same for converting mols Ca(OH)2 to mols Ca3(PO4)2.
2c. It is likely that the values in 2a and 2b will not agree which means one of them is not right. The correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is called the LR.

3. Now using the smaller value for mols,convert to grams. g = mols x molar mass.