Question

Consider the balanced equation below. How many grams of calcium phosphate are theoretically produced if we start with 3.40 moles of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) and 2.40 moles of \(\mathrm{Li}_{3} \mathrm{PO}_{4} ?\) Hint: Consider limiting reagent.

$$ 3 \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{Li}_{3} \mathrm{PO}_{4} \rightarrow 6 \mathrm{LiNO}_{3}+\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} $$

Answer 1

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

Answer 2

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