Consider the following reaction: 3 PbCl 2 + 2 NH 4 3 PO 4 → Pb 3 PO 4 2 + 6 NH 4 Cl If 72 mmol PbCl 2 is mixed with 50 mmol of NH 4 3 PO 4 , then amount of Pb 3 PO 4 2 formed in mmol . (nearest integer)

Step-by-step solution

The reaction is given as, 3 PbCl 2 + 2 NH 4 3 PO 4 → Pb 3 PO 4 2 + 6 NH 4 Cl The products formed in the above reaction is Pb 3 PO 4 2 and NH 4 Cl . According to the above stoichiometric equation, three mole of each lead chloride and two moles of NH 4 3 PO 4 is required to make one mole of Pb 3 PO 4 2 and 6 moles of NH 4 Cl Here the Limiting Reagent is PbCl 2 mmol of Pb 3 ( PO 4 ) 2 formed = mmol of PbCl 2 reacted 3 = 72 3 = 24 m mol