Stoichiometry
The calculation of the quantities of
chemical substances involved in
chemical reactions.
Limiting Reactants
Definition
A limiting reactant is that substance that causes a reaction to
stop before other reactants have combined to make new
substances.
When a reaction has a limiting reactant, one or more reactants
are left over after the limiting reactant has completely combined
to make new substances.
Limiting Reactants
Consider the following complex chemical reaction formula:
2 Graham Crackers + 4 Marshmallows + 3 Pieces Chocolate → 1 S’More
Chemicals on hand:
50 graham crackers, 80 marshmallows, and 65 pieces of chocolate
1. What is the limiting reactant? What will we run out of first?
2. How many S’Mores can you make?
3. If you borrowed 10 graham crackers, 20 marshmallows and 4 pieces of
chocolate from your neighbor and added them to yours, how many more
S’Mores could you make?
Limiting Reactants
2 Graham Crackers + 4 Marshmallows + 3 Pieces Chocolate → 1 S’More
50 graham crackers, 80 marshmallows, and 65 pieces of chocolate
1. What is the limiting reactant?
50 GC → 25 S’M
80 MM → 20 S’M
65 PC → 21.7 S’M
It is clear that Marshmallows is the
limiting reactant.
When all the marshmallows are used
up, you still have 10 graham crackers
and 5 pieces of chocolate left over.
Limiting Reactants
2 Graham Crackers + 4 Marshmallows + 3 Pieces Chocolate → 1 S’More
50 graham crackers, 80 marshmallows, and 65 pieces of chocolate
2. How many total S’Mores can you make with these chemicals??
50 GC → 25 S’M
80 MM → 20 S’M
65 PC → 31.7 S’M
20
Limiting Reactants
2 Graham Crackers + 4 Marshmallows + 3 Pieces Chocolate → 1 S’More
50 graham crackers, 80 marshmallows, and 65 pieces of chocolate
You have made 20 s’mores and have 10 GC and 5 PC left.
3. If you borrowed 10 graham crackers, 20 marshmallows and 4 pieces of
chocolate from your neighbor and added them to yours, how many more
S’Mores could you make?
More S’Mores
A. 10 GC + 10 GC =
20 GC
10
B.
0 MM + 20 MM = 20 MM
5
C.
5 PC + 4 PC =
3
9 PC
The limiting reactant NOW is chocolate, and you can make
3 more S’Mores, with 14 GC and 8 MM left over.
Review: Steps to Solving Chemical
Equation Word Problems
1. Write the equation using chemical formulae
2. Calculate 1 mole of each substance
3. Balance the equation
4. Find equivalent molar masses
5. Identify the knowns and unknowns
6. Set up and solve a ratio and proportion
Consider the following reaction:
C + O2 →
12 g + 32 g
→
CO2
44 g
If one combined 12 g of carbon with 32 g of oxygen, the result
would be 44 g of product. The oxygen would be completely
used up when all of the carbon was consumed. The reaction
would stop.
One mole of carbon (12 g) combines with one mole of oxygen
gas (32 g) to produce one mole of carbon dioxide (44 g). This
is a balanced chemical reaction.
Now consider:
C + O2 →
24 g + 32 g
CO2
→ 44 g
If one combined 24 g of carbon with 32 g of oxygen, the result
would be ? g of product.
The oxygen would be completely used up before all of the
carbon was consumed. The reaction would stop when the
oxygen was used up and there would be excess, unreacted,
carbon. The oxygen limits this reaction.
Conversely, if one started with 12 g of carbon and 64g of
oxygen, the reaction would stop when the carbon was used
up and there would be excess, unreacted, oxygen. The
reaction would still produce 44 g of product.
Steps to solving limiting reaction problems
Problem to solve:
Methane (carbon tetrahydride), a natural gas, is composed of carbon
and hydrogen, much like wood. It combines with oxygen to produce
carbon dioxide and water.
A. If we have 100 grams of methane and 100 grams of oxygen
and cause them to react, which is the limiting reactant and how much
of the other will remain after the reaction stops?
B. How much water is produced?
Step 1: Identify the reactants and products and then write the
chemical equation using appropriate chemical symbols and
formulas.
carbon tetrahydride + oxygen → carbon dioxide + water
CH4
+
O2
→
CO2
+
H2O
Step 2: Before balancing the equation, use your periodic table to
find the mass of one mole of each element or formula unit. Write
the molar masses above the element or formula unit.
1 mole =
16 g
CH4 +
32 g
O2
44 g
→
CO2
18g
+
H2O
(molar masses)
Step 3: Balance the equation using the correct coefficients.
16 g
CH4
32 g
+
2 O2
→
44 g
18g
CO2 +
2 H2O
Step 4: Multiply the molar masses by the coefficients to get the
balanced mass equivalents. Write this number below the element
or formula unit.
16 g
CH4
16 g
32 g
+
2 O2
64 g
→
44 g
18g
CO2 +
2 H2O
44 g
36 g
(molar masses)
At this point, the procedure changes. We will need to solve two
problems, one using each known quantity to see how much of the
other is needed.
Step 5: Using your problem statement and the balanced mass
equivalents, set up two ratios to solve for your unknown.
5 A. First ratio: Start with 100 grams of methane and see
how much oxygen is needed.
CH4 +
16 g
100 g
Solving for X:
2 O2 → CO2 + 2 H2O
64 g
44 g
36 g
X
16 X = 6400 g
X = 400 g (of oxygen)
Conclusion: If we started with 100 g of methane, we would need
400 g of oxygen for the whole reaction. (We started with 100 g O2)
5 B. Second ratio: Start with 100 g of oxygen and see how
much methane is needed.
CH4 +
2 O2
16 g
X
64 g
100g
Solving for X:
→ CO2 + 2 H2O
44 g
64 X = 1600 g
36 g
X = 25 g (of methane)
Conclusion: If we started with 100 g of oxygen, it would be used up
when 25 g of methane had been combined. (We started with 100 g of
methane.)
Step 6: Analysis. Compare the results to determine the limiting
reactant.
Step 6 A. In order to burn 100 g of methane, 400 g of
oxygen would be needed. It was not available, though, because
we only started with 100 g of oxygen.
Step 6 B. When 100 g of oxygen is used up, the reaction
stops, but only 25 g of methane reacted, and 75 g of methane is
left over, or is in excess.
Conclusion: Since the reaction stopped when the oxygen was
used up, and there was still 75 g of methane to burn, then the
oxygen is the limiting reactant. The reaction stops (is limited)
because there is no more oxygen to react.
Step 7: Solving for quantity of water produced.
Since we now know that 100 g of oxygen is the limiting reactant,
we can use this information to calculate the amount of water
produced:
CH4 +
2 O2
16 g
64 g
100g
Solving for X:
→ CO2 + 2 H2O
44 g
64 X = 100 x 36 g
36 g
X
X = 56.25 g
We can check this answer. Since we also know that 25 g of
methane was used to combine with 100 g of oxygen, we can also
substitute and use the methane values.
CH4 +
2 O2
16 g
25 g
64 g
→ CO2 + 2 H2O
44 g
Again, solving for X:
Surprise of all surprises,
36 g
X
16 X = 36 x 25 g
X = 56.25 g
Practice Problem #1
Hydrogen gas (diatomic) combines with oxygen gas (diatomic) to
make water. If you start with 10 grams of hydrogen gas and 40
grams of oxygen:
A. What is the limiting reactant?
B. How much water will be produced?
Practice Problem #1
Hydrogen gas (diatomic) combines with oxygen gas (diatomic) to
make water. If you start with 10 grams of hydrogen gas and 40
grams of oxygen:
Step 1: Write the chemical equation:
H2
+
O2
→
H2O
Practice Problem #1
Hydrogen gas (diatomic) combines with oxygen gas (diatomic) to
make water. If you start with 10 grams of hydrogen gas and 40
grams of oxygen:
Step 2: Find one mole of each substance:
1 mole = 2 g
H2
32 g
+
O2
18 g
→
H2O
Practice Problem #1
Hydrogen gas (diatomic) combines with oxygen gas (diatomic) to
make water. If you start with 10 grams of hydrogen gas and 40
grams of oxygen:
Step 3: Balance the equation:
1 mole = 2 g
2 H2
32 g
+
O2
18 g
→
2 H 2O
Practice Problem #1
Hydrogen gas (diatomic) combines with oxygen gas (diatomic) to
make water. If you start with 10 grams of hydrogen gas and 40
grams of oxygen:
Step 4: Calculate balanced molar masses:
1 mole = 2 g
2 H2
4g
32 g
+
O2
32 g
18 g
→
2 H2O
36 g
Practice Problem #1
Hydrogen gas (diatomic) combines with oxygen gas (diatomic) to
make water. If you start with 10 grams of hydrogen gas and 40
grams of oxygen:
Step 5: Find oxygen needed for 10 g hydrogen and find
hydrogen needed for 40 g oxygen:
2 H2
4g
4 g = 32 g
10 g
X
Oxygen
+
O2
32 g
X = 80 g
→
2 H2O
36 g
4 g = 32 g
X
40 g
X=5g
Hydrogen
Practice Problem #1
Hydrogen gas (diatomic) combines with oxygen gas (diatomic) to
make water. If you start with 10 grams of hydrogen gas and 40
grams of oxygen:
Step 6: Analysis
2 H2
4g
4 g = 32 g
10 g
X
+
O2
32 g
X = 80 g
→
2 H2O
36 g
4 g = 32 g
X
40 g
X=5g
The reaction stops when 40 g oxygen are used, so oxygen is
the limiting reactant.
Practice Problem #1
Hydrogen gas (diatomic) combines with oxygen gas (diatomic) to
make water. If you start with 10 grams of hydrogen gas and 40
grams of oxygen:
Step 6 extension: Use limiting reactant to solve for
product.
2 H2
4g
+
O2
32 g
32 g = 36 g
40 g
X
→
2 H2O
36 g
X = 45 g water
Homework Problem # 1.
Hydrogen phosphate, when combined with potassium hydroxide
produces potassium phosphate and water. You start with 20.0 g
of hydrogen phosphate and 30.0 g of potassium hydroxide.
A. What is the limiting reactant?
B. How much of each product will be produced?
C. How much of the excess reagent is left over?
Homework Problem # 1.
Hydrogen phosphate, when combined with potassium hydroxide
produces potassium phosphate and water. You start with 20.0 g
of hydrogen phosphate and 30.0 g of potassium hydroxide.
Step 1. Write a chemical equation:
H3PO4
+
KOH
→
K3PO4
+
H2O
Homework Problem # 1.
Hydrogen phosphate, when combined with potassium hydroxide
produces potassium phosphate and water. You start with 20.0 g
of hydrogen phosphate and 30.0 g of potassium hydroxide.
Step 2. Determine 1 mole of each substance:
1 mole:
98 g
H3PO4
56 g
+
KOH
212 g
→
K3PO4
18 g
+
H2O
Homework Problem # 1.
Hydrogen phosphate, when combined with potassium hydroxide
produces potassium phosphate and water. You start with 20.0 g
of hydrogen phosphate and 30.0 g of potassium hydroxide.
Step 3. Balance the equation:
1 mole:
98 g
H3PO4
56 g
+ 3 KOH
212 g
→
K3PO4
18 g
+
3 H2 O
Homework Problem # 1.
Hydrogen phosphate, when combined with potassium hydroxide
produces potassium phosphate and water. You start with 20.0 g
of hydrogen phosphate and 30.0 g of potassium hydroxide.
Step 4. Calculate equivalent molar masses:
1 mole:
98 g
H3PO4
98 g
56 g
+
3 KOH
168 g
212 g
→
K3PO4
212 g
18 g
+
3 H2 O
54 g
Homework Problem # 1.
Hydrogen phosphate, when combined with potassium hydroxide
produces potassium phosphate and water. You start with 20.0 g
of hydrogen phosphate and 30.0 g of potassium hydroxide.
Step 5. Use known information to calculate reactant requirements:
1 mole:
98 g
H3PO4
56 g
+
98 g
For 20.0 g H3PO4:
98 g = 168 g
20 g
X
X = 34.3 KOH
3 KOH
212 g
→
K3PO4
168 g
18 g
+
212 g
3 H2 O
54 g
For 30.0 g KOH:
98 g = 168 g
X
30 g
X = 17.5 g H3PO4
Homework Problem # 1.
Hydrogen phosphate, when combined with potassium hydroxide
produces potassium phosphate and water. You start with 20.0 g
of hydrogen phosphate and 30.0 g of potassium hydroxide.
Step 6. Analysis:
H3PO4
+
3 KOH
For 20.0 g H3PO4:
98 g = 168 g
20 g
X
X = 34.3 KOH
We need 34.3 g of KOH to
completely react with 20.0 g of
H3PO4. We only have 30.0 g
→
K3PO4
+
3 H2 O
For 30.0 g KOH:
98 g = 168 g
X
X = 17.5 g H3PO4
30 g
We need 17.5 g of H3PO4 to react
with 30.0 g of KOH. We have
MORE than enough H3PO4. KOH is
the limiting reagent.
Homework Problem # 1.
Hydrogen phosphate, when combined with potassium hydroxide
produces potassium phosphate and water. You start with 20.0 g
of hydrogen phosphate and 30.0 g of potassium hydroxide.
Extension: Use the limiting reactant to solve for products:
H3PO4
98g
+
3 KOH
168 g
K3PO4
+
212 g
3 H2O
54 g
For H2O :
For K3PO4:
168 g = 212 g
30 g
X
→
X = 37.9 g K3PO4
168 g = 54 g
30 g
X
X = 9.6 g H2O
Homework Problem # 1.
Hydrogen phosphate, when combined with potassium hydroxide
produces potassium phosphate and water. You start with 20.0 g
of hydrogen phosphate and 30.0 g of potassium hydroxide.
Extension: Use the limiting reactant to determine excess
reactant.
H3PO4
98g
+
3 KOH
168 g
→
K3PO4
212 g
Since KOH is the Limiting Reactant,
solve for required H3PO4:
Start with:
Need:
168 g = 98 g
30 g
X
Excess:
X = 17.5 g H3PO4
+
3 H2O
54 g
20.0 g H3PO4
17.5 g
2.5 g H3PO4