1. Chapter 7 Chemical Quantities in Reactions
7.2
Limiting Reactants

2. Limiting Reactant A limiting reactant in a chemical reaction is the
substance that
Is used up first.
Stops the reaction.
Limits the amount of product that can form.

3. Reacting Amounts In a table setting, there is 1
plate, 1 fork, 1 knife, and
1 spoon.
How many table settings are
possible from 5 plates, 6 forks,
4 spoons, and 7 knives?
What is the limiting item?
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4. Reacting Amounts Four table settings can be made.
Initially Use Left over
plates
forks
spoons
knives
The limiting item is the spoon.

5. Example of Everyday Limiting Reactant
How many peanut butter sandwiches could be made from 8 slices bread and 1 jar of peanut butter?
With 8 slices of bread, only 4 sandwiches could be made. The bread is the limiting item.
Copyright © by Pearson Education, Inc. publishing as Benjamin Cummings

6. Example of Everyday Limiting Reactant
How many peanut butter sandwiches could be made from 8 slices bread and 1 tablespoon of peanut butter?
With 1 tablespoon of peanut butter, only 1 sandwich could be made. The peanut butter is the limiting item.
Copyright © by Pearson Education, Inc. publishing as Benjamin Cummings

7. Limiting Reactants When 4.00 mol H2 is mixed with 2.00 mol Cl2,how
many moles of HCl can form?
H2(g) Cl2(g)  2HCl (g)
4.00 mol mol ??? mol
Calculate the moles of product from each reactant, H2 and Cl2.
The limiting reactant is the one that produces the smaller amount of product.

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Copyright © by Pearson Education, Inc. publishing as Benjamin Cummings

9. Limiting Reactants Using Moles
HCl from H2
4.00 mol H2 x 2 mol HCl = mol HCl
1 mol H (not possible)
HCl from Cl2
2.00 mol Cl2 x 2 mol HCl = mol HCl
1 mol Cl (smaller number)
The limiting reactant is Cl2 because it is used up first. Thus Cl2 produces the smaller number of moles of HCl.

10. Checking Calculations
Initially H2
4.00 mol
Cl2
2.00 mol
2HCl
0 mol
Reacted/
Formed
-2.00 mol
+4.00 mol
Left after reaction
Excess
0 mol
Limiting

11. Limiting Reactants Using Mass
If 4.80 mol Ca mixed with 2.00 mol N2, which is the
limiting reactant? 3Ca(s) + N2(g)  Ca3N2(s)
Moles of Ca3N2 from Ca
4.80 mol Ca x 1 mol Ca3N2 = 1.60 mol Ca3N mol Ca (Ca used up)
Moles of Ca3N2 from N2
2.00 mol N2 x 1 mol Ca3N2 = mol Ca3N2
1 mol N (not possible)
All Ca is used up when 1.60 mol Ca3N2 forms. Thus, Ca is the limiting reactant and N2 is our excess reactant.

12. Limiting Reactants Using Mass
Calculate the mass of water produced when
8.00 g H2 and 24.0 g O2 react?
2H2(g) + O2(g) H2O(l)
a) Determine the limiting and excess reactants.

13. Limiting Reactants Using Mass
Calculate the grams of H2 for each reactant.
H2:
8.00 g H2 x 1 mol H2 x 2 mol H2O x g H2O
2.016 g H mol H mol H2O
= 71.5 g H2O (not possible)
O2:
24.0 g O2 x 1 mol O2 x 2 mol H2O x g H2O
32.00 g O mol O mol H2O
= 27.0 g H2O (smaller)
O2 is the limiting reactant
H2 is the excess reactant

14. Extension Problem b) How much excess reactant remains?
Always start all calculations from the limiting reactant:
24.0 g O2 x 1 mol O2 x 2 mol H2 x g H2 = 3.03 g
32.00 g O mol O mol H2
This number indicates how much H2 was reacted. So to determine how
much excess reactant remains you must subtract from the starting value.
Excess Remaining = 8.00 g of H g of H2
= 4.97 g of H2