Limiting Reagent Stoichiometry | The Basic Ratio Trick
One of the most frequent hurdles in the Mole Concept & Stoichiometry chapter is identifying the Limiting Reagent (LR). If two reactants are mixed, which one runs out first? If you pick the wrong one, your entire product calculation will be wrong.
The Question: If 72 mmol of PbCl2 and 50 mmol of (NH4)3PO4 are mixed, find out the mmol of Pb3(PO4)2 formed.
Let's use a brilliant tabular shortcut to instantly identify the limiting reagent and find the product yield without getting tangled in fractions.
Video Tutorial: The Ratio Multiplier Method
Watch Abhishek Sengar sir from CHEMCA expertly write the balanced equation, set up the basic ratio table, and deduce the exact amount of product formed.
Step 1: Write and Balance the Equation
Before any math can happen, we must ensure the chemical equation is perfectly balanced. If the coefficients are wrong, the basic ratio is wrong.
Step 2: Establish the Basic Ratio
The coefficients from the balanced equation give us the fundamental "recipe" or Basic Ratio for the reaction. It tells us exactly how many parts of each molecule are required and produced.
- For every 3 moles of PbCl2 used...
- We require exactly 2 moles of (NH4)3PO4.
- This produces 1 mole of Pb3(PO4)2.
- And produces 6 moles of NH4Cl.
Step 3: Identify the Limiting Reagent (LR)
We are given 72 mmol of PbCl2 and 50 mmol of (NH4)3PO4. Let's see which one runs out first by applying a simple multiplier.
Look at the basic ratio for PbCl2, which is 3. We have 72 mmol. How do we get from 3 to 72? We multiply by 24!
If we multiply the entire Basic Ratio row by 24, we will instantly see exactly how much of every other chemical is needed to fully react all 72 mmol of PbCl2.
| State | 3 PbCl2 | + 2 (NH4)3PO4 | → 1 Pb3(PO4)2 | + 6 NH4Cl |
|---|---|---|---|---|
| Basic Ratio | 3 | 2 | 1 | 6 |
| Multiply by 24 | 72 | 48 | 24 | 144 |
| Given Amount | 72 mmol | 50 mmol | - | - |
The table reveals everything! To fully consume the 72 mmol of PbCl2, we only need 48 mmol of (NH4)3PO4. Since we were given 50 mmol, we have more than enough.
Because (NH4)3PO4 is left over, it is the Excess Reagent. Therefore, PbCl2 is the Limiting Reagent. All calculations for products must be based strictly on the limiting reagent.
Fig: Multiplying the entire basic ratio row by a single factor eliminates the need for cross-multiplication or fractions!
Step 4: Final Answer
Look at the product column for Pb3(PO4)2 in our table. The basic ratio is 1. Multiplying by our factor of 24 gives us exactly 24 mmol.
Practice Questions for JEE & NEET
Let's test your ability to extract more information from this exact same table!
Question 1: After the reaction goes to 100% completion, how many millimoles (mmol) of the excess reagent will be left unreacted in the beaker?
Answer: 2 mmol of (NH4)3PO4 will remain.
Reasoning:
The table makes this incredibly easy.
1. We were initially given 50 mmol of (NH4)3PO4.
2. The basic ratio table showed that exactly 48 mmol of it was required to fully consume the limiting reagent (PbCl2).
3. Unreacted Amount = Initial Amount - Reacted Amount = 50 - 48 = 2 mmol.
The reaction stops because there is no more PbCl2 available to react with these final 2 mmol!
Question 2: Fundamentally, why is it mathematically incorrect to calculate the amount of product formed using the starting amount of the excess reagent?
Answer: Because the reaction physically stops before the excess reagent is fully consumed.
Reasoning:
Stoichiometry calculates theoretical yields based on the assumption that the chemical you are doing math with completely reacts.
If you used 50 mmol of (NH4)3PO4 as the basis for your math, you would calculate a product yield that implies all 50 mmol turned into product. However, as we saw in Question 1, only 48 mmol actually turned into product. The remaining 2 mmol did absolutely nothing! Therefore, using the starting amount of the excess reagent will always yield a falsely high (and incorrect) product calculation.
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