1. Stoichiometry, The Mole,
and Quantitative
Review our discussion of
Chemistry mathematical applications to
chemistry: The mole and
Dimensional analysis.
• How many atoms are
in 12.00 grams of Iron?
Nylon Rope

2. We tie together the concepts of Formula
writing, identifying reactions, and
quantitative chemistry
If we take a look again at the reaction
from the Types of Chemical Reactions
chapter between Cl2 and NaBr, this
time from a quantitative standpoint.
Cl2 + 2 NaBr Br2 + 2 NaCl
Polypropylene knife handles

3. A Sample Problem...
Cl2
+
2 NaBr
Br2
+
2 NaCl
If we react 5.00 grams of NaBr with excess Cl2, how many grams of Br2
will be formed?
Construct a pathway…
Grams Reactant-->Moles Reactant-->Moles Product-->Grams Product
The Solution...
5.00g NaBr x 1 mole NaBr/102.90 g NaBr x 1 mole Br2/2 mole NaBr x
159.80g Br2/1 mole Br2=
3.88 g Br2

4. Try This One...
CaCl2
+
Na2CO3
CaCO3
+ 2 NaCl
• How many grams of CaCO3 will be formed if 8.00
grams of CaCl2 are completely reacted with
Na2CO3?
Limestone formation
in South Africa

5. Limiting Reactant Problems
Ideally, each reactant in a reaction is
completely used up. Sometimes, however,
one reactant is used up before the other
and the reaction stops. This is called the
Limiting Reactant.
Let’s Take Another Look
Cl2
+
2 NaBr
Br2
+
2 NaCl

6. The Solution...
If we reacted 5.00 g of NaBr with 8.00 g of Cl2, how many grams of Br2 will be formed?
What is the limiting reactant? How many grams of excess reactant remain?
5.00g NaBr x 1 mole NaBr/102.90g NaBr x 1 mole Br2/2 mole NaBr x 159.80g Br2/1mole
Br2=
3.88g Br2
*NaBr is the Limiting Reactant
8.00g Cl2 x 1 mole Cl2/70.90g Cl2 x 1 mole Cl2/1mole Br2 x 159.80g Br2/1 mole Br2=
18.03 g Br2
*Cl2 is the excess reactant
3.88g Br2 x 1mole Br2/159.80 g Br2 x 1 mole Cl2/1mole Br2 x 70.90 g Cl2/1mole Cl2=
1.72 g Cl2 needed