To solve this problem, we will use the stoichiometry of the balanced chemical reaction. The balanced equation shows that 3 moles of PbCl2 react with 2 moles of (NH4)3PO4 to produce 1 mole of Pb3(PO4)2.
The reaction is:
3PbCl2+2(NH4)3PO4→Pb3(PO4)2+6NH4Cl
The molar ratio of PbCl2 to Pb3(PO4)2 is 3:1, and the molar ratio of (NH4)3PO4 to Pb3(PO4)2 is 2:1. We need to determine which reactant is the limiting reagent because it will dictate the amount of Pb3(PO4)2 produced.
The stoichiometric calculations are as follows:
For PbCl2:
Moles of Pb3(PO4)2 formed from PbCl2=3 mmol of PbCl2/mmol of Pb3(PO4)272 mmol of PbCl2=24 mmol
For (NH4)3PO4:
Moles of Pb3(PO4)2 formed from (NH4)3PO4=2 mmol of (NH4)3PO4/mmol of Pb3(PO4)250 mmol of (NH4)3PO4=25 mmol
Now we can identify the limiting reagent by comparing the two amounts of Pb3(PO4)2 that could be produced. The smaller quantity will be the actual amount produced since the limiting reagent restricts the reaction.
Since the PbCl2 can produce only 24 mmol of Pb3(PO4)2 versus the 25 mmol that could be produced by (NH4)3PO4, PbCl2 is the limiting reagent.
Therefore, the amount of Pb3(PO4)2 formed is 24 mmol (as a nearest integer).
