Molarity is moles per liter, so moles equals concentration times volume.
Molarity Moles of solute per liter of solution.
Step 1: Something to Notice
Watch first. The explanation comes later.
A reaction is run with more of reactant A than of reactant B, and A runs out first.
How can the larger amount be the one exhausted?
Step 2: Find Out
Isolate the supplied ratio and note which reactant limits the product each time.
Step 1 — Predict
Is the reactant present in the smaller amount always the limiting one?
Choose what you think will happen. You cannot see the experiment until you do — guessing first is what makes it worth watching.
Step 3: So Here Is Why
Now the explanation, after you have seen it happen.
Dilution conserves the solute
C₁V₁ = C₂V₂ states that the moles did not change.
Divide each amount by its coefficient
The smallest result is the limiting reagent, whatever the raw amounts.
Limiting reagent The reactant that runs out first, setting the maximum yield.
The limiting reagent gives the theoretical yield
It is the maximum obtainable, before any practical loss.
Yields fall short for identifiable reasons
Equilibrium, side reactions and transfer losses — and above 100% means impure.
Concentration times volume is a number of moles
Molarity is moles per liter, so multiplying by liters gives moles. That single relationship is what connects solution measurements to reaction stoichiometry.
Solutions have limiting reactants too
Mixing two solutions requires converting each to moles and comparing against the equation ratio. Volume alone says nothing about which will run out first.
Step 4: A Common Mistake
Lots of people think
“The reactant present in the smaller amount is the limiting one.”
Step 5: Why an Industrial Process Runs a Reactant in Excess Deliberately
The same idea somewhere new.
A plant will often supply one reactant far in excess of the stoichiometric requirement, which looks wasteful and is not. Excess of a cheap reactant drives an equilibrium towards products, raising the conversion of the expensive one, and the excess is usually recovered and recycled. Which reactant to make limiting is therefore an economic decision as much as a chemical one: the expensive, hazardous or hard-to-separate material is the one to consume completely. It is a good illustration that a stoichiometric calculation tells you what is possible and not what is sensible.
Step 6: Think It Through
Practice makes it stick.
The Larger Amount Runs Out
Problem 1 of 2
More of A than B is supplied and A runs out first. How?
A Yield Above 100%
Problem 2 of 2
A student reports a 108% yield. What does that indicate?
Moles in Solution
1 of 5
What is molarity?
2 of 5
What does C₁V₁ = C₂V₂ express?
3 of 5
How is the limiting reagent identified?
4 of 5
Why do actual yields fall short?
5 of 5
Why might a process run one reactant in excess?
Step 7: Quick Check
Show what you know.
Question 1 of 1
Is the reactant present in the smaller amount the limiting one?
Step 8: Explain It in Writing
Claim, evidence, then reasoning.
The question
Explain how to identify a limiting reagent and calculate a percentage yield in solution.
Fill in all three boxes. The reasoning box is the one that matters most.
What You Found Out
- Moles equals concentration times volume, which is all of solution stoichiometry.
- Dilution conserves solute, which is what C₁V₁ = C₂V₂ states.
- The limiting reagent is found by dividing each amount by its coefficient.
- Yields fall short for identifiable reasons, and above 100% means impurity.