How many grams of calcium phosphate are theoretically produced ifwe start with 3.40 moles of...

First we must get the balanced equation

2.4 mol of ?

I will write the net ionic

3 Ca2+ + 2 PO43- = Ca3(PO4)2

the ratio between Ca2+ and PO43- is 3 : 2

moles PO43- required = 3.40 x 2 /3 = 2.27 so PO43- is the limiting reactant

the ratio between PO43- and Ca3(PO4)2 is 2 : 1

moles Ca3(PO4)2 = 2.27 x 1 /2 = 1.13

mass calcium phosphate = 1.13 mol x 310.18 g/mol=350.5 g

3 Ca(NO3)2 + 2 Li3PO4 >> 6 LiNO3 + Ca3(PO4)2

the ratio between Ca(NO3)2 and Li3PO4 is 3 :2

3 : 2 = 3.40 : x

x = 2.26 moles Li3PO4 needed for the reaction

we have 2.40 moles Li3PO4 so it is in excess

Ca(NO3)2 limiting reactant

The ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 : 1

3 : 1 = 3.4 : x

x = 1.13 moles Ca3(PO4)2 produced

Molecular mass Ca3(PO4)2 = 310.18 g/mol

grams = 310.18 g/mol x 3.40 moles = 10546