Limiting Reactant Notes
(a continuation of Stoichiometry…)
The Idea
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In every chemical reaction, there is one reactant that will be run out
(This is called the ______________________________)
This will limit the amount of product that can form.
The reaction will stop at that point.
There are then also ____________________________.
Calculating the Limiting Reactant (LR)
Sample Problem: S8 + 4 Cl2  4 S2Cl2
If you have 200.0 g S8 and 100.0 g Cl2, what is the limiting reactant?
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Step 1: Find the moles of each reactant (this is a g  mol calculation)
Calculations:
200.0 g S8
1
100.0 g Cl2
1
NOTE: If moles are given, Step 1 is already done!
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Step 2: Compare Mole Ratios
A)
What is the _____________________ ratio from the balanced chemical equation?
S8 + 4 Cl2  4 S2Cl2
Mol S8:
Mol Cl2:
Ratio:
B) What is the mole to mole ratio calculated from the starting conditions?
Mol S8:
Mol Cl2:
Calculating the Ratio:
Is this enough? ____________
Therefore, ___________ is the limited reactant.
Calculating the Amount of Product Formed
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Complete a ______________________________ calculation using the LR
In the example: limiting reactant is Cl2 = _______________
The calculation:
Analyzing the Excess Reactant
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Excess that Reacted
Complete a mole to gram calculation using the limited and excess reactants
In our example:
o Limiting Reactant:
o Excess Reactant:
The Calculation:
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Excess that Remains
Subtract amount that reacted from the amount that you started with
In example:
o Amount Reacted =
o Amount Starting With =
The Answer: