Consider the following reaction:\[3\text{PbCl}_2 + 2(\text{NH}_4)_3\text{PO}_4 \rightarrow \text{Pb}_3(\text{PO}_4)_2 + 6\text{NH}_4\text{Cl}\]If 72 mmol of PbCl\(_2\) is mixed with 50 mmol of (\(\text{NH}_4\))\(_3\)PO\(_4\), then the amount of \(\text{Pb}_3(\text{PO}_4)_2\) formed is ______ mmol (nearest integer).
Correct Answer: 24
Approach Solution - 1
Limiting reagent is PbCl2.
Amount of Pb3(PO4)2 formed:
\( \text{mmol of Pb}_3(\text{PO}_4)_2 = \frac{\text{mmol of PbCl}_2 \text{ reacted}}{3} = 24 \text{ mmol} \)
Approach Solution -2
This is a stoichiometry problem that involves determining the amount of product formed from a given amount of reactants. The key is to first identify the limiting reactant, which will dictate the maximum amount of product that can be formed.
Concept Used:
The concept of a limiting reactant (or limiting reagent) is used in stoichiometry. When reactants are not mixed in the exact stoichiometric ratio as given by the balanced chemical equation, one reactant will be completely consumed before the others. This reactant is the limiting reactant, and it determines the theoretical yield of the product.
The steps to find the limiting reactant and the amount of product formed are:
- Write the balanced chemical equation for the reaction.
- Convert the given amounts of reactants into moles (or millimoles, as given).
- Calculate the amount of product that could be formed from each reactant, assuming the other reactant is in excess.
- The reactant that produces the smallest amount of product is the limiting reactant.
- The smallest amount of product calculated is the actual amount that will be formed.
Step-by-Step Solution:
Step 1: Write down the balanced chemical equation and the initial amounts of reactants.
The balanced equation is:
\[ 3\text{PbCl}_2 + 2(\text{NH}_4)_3\text{PO}_4 \rightarrow \text{Pb}_3(\text{PO}_4)_2 + 6\text{NH}_4\text{Cl} \]Initial amounts given:
- Amount of PbCl₂ = 72 mmol
- Amount of (NH₄)₃PO₄ = 50 mmol
Step 2: Determine the limiting reactant by calculating the amount of product, Pb₃(PO₄)₂, that can be formed from each reactant.
We will calculate the moles of Pb₃(PO₄)₂ formed from each reactant individually.
Step 3: Calculate the amount of Pb₃(PO₄)₂ formed from PbCl₂.
From the stoichiometry of the reaction, 3 moles of PbCl₂ produce 1 mole of Pb₃(PO₄)₂.
\[ \text{Amount of } \text{Pb}_3(\text{PO}_4)_2 = 72 \text{ mmol PbCl}_2 \times \frac{1 \text{ mmol } \text{Pb}_3(\text{PO}_4)_2}{3 \text{ mmol PbCl}_2} \] \[ = \frac{72}{3} = 24 \text{ mmol of } \text{Pb}_3(\text{PO}_4)_2 \]Step 4: Calculate the amount of Pb₃(PO₄)₂ formed from (NH₄)₃PO₄.
From the stoichiometry of the reaction, 2 moles of (NH₄)₃PO₄ produce 1 mole of Pb₃(PO₄)₂.
\[ \text{Amount of } \text{Pb}_3(\text{PO}_4)_2 = 50 \text{ mmol } (\text{NH}_4)_3\text{PO}_4 \times \frac{1 \text{ mmol } \text{Pb}_3(\text{PO}_4)_2}{2 \text{ mmol } (\text{NH}_4)_3\text{PO}_4} \] \[ = \frac{50}{2} = 25 \text{ mmol of } \text{Pb}_3(\text{PO}_4)_2 \]Final Computation & Result:
Step 5: Compare the amounts of product and identify the limiting reactant.
The amount of Pb₃(PO₄)₂ that can be formed is 24 mmol (from PbCl₂) and 25 mmol (from (NH₄)₃PO₄).
Since the amount of product formed from PbCl₂ (24 mmol) is less than the amount that could be formed from (NH₄)₃PO₄ (25 mmol), PbCl₂ is the limiting reactant. The reaction will stop once all the PbCl₂ is consumed.
Therefore, the maximum amount of Pb₃(PO₄)₂ that can be formed is determined by the limiting reactant.
Amount of Pb₃(PO₄)₂ formed = 24 mmol.
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