How many grams of calcium phosphate (precipitate) could be produced by combining 12.096g of calcium nitrate dissolved in water with 16.498g of rubidium phosphate dissolved in water? Report your answer to 3 decimal places.

Answer 1

This is a limiting reagent problem. I know that because amounts for BOTH reactants are given.

Write and balance the equation.

Convert Ca(NO3)2 to mols. mols = grams/m
molar mass.
Do the same for Rb3PO4.

Using the coefficients in the balanced equation,convert mols of each of the reactants to mols of Ca3(PO4)2. It is likely you will get two values; one of them must be wrong. The correct answer in limiting reagent problems is ALWAYS the smaller one and the reagent responsible for that smaller number is the limiting reagent.

Now convert the smaller number to grams. g = mols x molar mass.

Answer 2

Can you explain how to convert mols of each of the reactants to mols of Ca3(PO4)2 more please?

Answer 3

3Ca(NO3)2 + 2Rb3PO4 ==> Ca3(PO4)2 + 6RbNO3
mols Ca(NO3) = about 12.096/164.09 = about 0.0737.

It's a factor. Notice how the units you don't want cancel and the unit you want to convert to remains.
0.0737 mols Ca(NO3)2 x [1 mol Ca3(PO4)3/3 mols Ca(NO3)2] = 0.0737 x (1/3) = ? mols Ca3(PO4)2
Using factor like this you can convert anything to anything. The factors are
[what you have x (mols what you want/mols what you have)] where mols of each are the coefficients in the balanced equation.