Consider the following reaction: 3PbCl_2 + 2(NH_4)_3PO_4 rightarrow Pb_3(PO_4)_2 + 6NH_4Cl$$3PbCl_2 + 2(NH_4)_3PO_4 \rightarrow Pb_3(PO_4)_2 + 6NH_4Cl$$ If 72 mathrm~mmol$72 \mathrm{~mmol}$ of PbCl_2$PbCl_2$ is mixed with 50 mathrm~mmol$50 \mathrm{~mmol}$ of (NH_4)_3PO_4$(NH_4)_3PO_4$, then amount of Pb_3(PO_4)_2$Pb_3(PO_4)_2$ formed is ... mmol. (nearest integer)

### Related Formula textMoles of Product = textMoles of Limiting Reagent times fractextStoichiometry of ProducttextStoichiometry of Limiting Reagent $$ \text{Moles of Product} = \text{Moles of Limiting Reagent} \times \frac{\text{Stoichiometry of Product}}{\text{Stoichiometry of Limiting Reagent}} $$ ### Core Logic From the balanced chemical equation: 3 text moles of PbCl_2 text react with 2 text moles of (NH_4)_3PO_4$3 \text{ moles of } PbCl_2 \text{ react with } 2 \text{ moles of } (NH_4)_3PO_4$. Let's find the limiting reagent (L.R.) by dividing given millimoles by stoichiometric coefficients: For PbCl_2$PbCl_2$: frac723 = 24$\frac{72}{3} = 24$ For (NH_4)_3PO_4$(NH_4)_3PO_4$: frac502 = 25$\frac{50}{2} = 25$ Since 24 < 25$24 < 25$, PbCl_2$PbCl_2$ is the limiting reagent and will completely consume. ### Step 1: Calculate Product Moles Moles of Pb_3(PO_4)_2$Pb_3(PO_4)_2$ formed depends entirely on PbCl_2$PbCl_2$. 3 mmol of PbCl_2$PbCl_2$ produces 1 mmol of Pb_3(PO_4)_2$Pb_3(PO_4)_2$. Therefore, 72 mmol of PbCl_2$PbCl_2$ will produce: frac13 times 72 = 24 mathrm~mmol$\frac{1}{3} \times 72 = 24 \mathrm{~mmol}$ of Pb_3(PO_4)_2$Pb_3(PO_4)_2$. ### Pattern Recognition Always identify the Limiting Reagent by taking the ratio n / textcoefficient$n / \text{coefficient}$. The smallest ratio dictates the extent of the reaction. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry