stoichiometry the calculation of the quantities of chemical substances involved in chemical...

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



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. 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



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?

A. 10 GC + 10 GC = 20 GC

B. 0 MM + 20 MM = 20 MM

C. 5 PC + 4 PC = 9 PC

More S’Mores

10

5

3

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 → CO2

12 g + 32 g → 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 → CO2

24 g + 32 g → 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.

44

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 32 g 44 g 18g (molar masses)

CH4 + O2 → CO2 + H2O

Step 3: Balance the equation using the correct coefficients.

16 g 32 g 44 g 18g

CH4 + 2 O2 → 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 32 g 44 g 18g (molar masses)

CH4 + 2 O2 → CO2 + 2 H2O

16 g 64 g 44 g 36 g

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 + 2 O2 → CO2 + 2 H2O16 g 64 g 44 g 36 g

100 g X

Solving for 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 → CO2 + 2 H2O

16 g 64 g 44 g 36 g X 100g

Solving for X: 64 X = 1600 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 → CO2 + 2 H2O

16 g 64 g 44 g 36 g 100g X

Solving for X: 64 X = 100 x 36 g 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 → CO2 + 2 H2O

16 g 64 g 44 g 36 g 25 g X Again, solving for X: 16 X = 36 x 25 g

Surprise of all surprises, 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


Step 1: Write the chemical equation:

H2 + O2 → H2O

Practice Problem #1


Step 2: Find one mole of each substance:

H2 + O2 → H2O

1 mole = 2 g 32 g 18 g

Practice Problem #1


Step 3: Balance the equation:

H2 + O2 → H2O

1 mole = 2 g 32 g 18 g

2 2

Practice Problem #1


Step 4: Calculate balanced molar masses:

2 H2 + O2 → 2 H2O

1 mole = 2 g 32 g 18 g

4 g 32 g 36 g

Practice Problem #1


Step 5: Find oxygen needed for 10 g hydrogen and find hydrogen needed for 40 g oxygen:

2 H2 + O2 → 2 H2O

4 g 32 g 36 g

4 g = 32 g X = 80 g10 g X

Oxygen

4 g = 32 g X = 5 g X 40 g

Hydrogen

Practice Problem #1


Step 6: Analysis

2 H2 + O2 → 2 H2O

4 g 32 g 36 g

4 g = 32 g X = 80 g10 g X

4 g = 32 g X = 5 g X 40 g

The reaction stops when 40 g oxygen are used, so oxygen is the limiting reactant.

Practice Problem #1


Step 6 extension: Use limiting reactant to solve for product.

2 H2 + O2 → 2 H2O

4 g 32 g 36 g

32 g = 36 g X = 45 g water40 g X

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?



Step 1. Write a chemical equation:

H3PO4 + KOH → K3PO4 + H2O



Step 2. Determine 1 mole of each substance:

1 mole: 98 g 56 g 212 g 18 g

H3PO4 + KOH → K3PO4 + H2O



Step 3. Balance the equation:

1 mole: 98 g 56 g 212 g 18 g

H3PO4 + KOH → K3PO4 + H2O3 3



Step 4. Calculate equivalent molar masses:

1 mole: 98 g 56 g 212 g 18 g

H3PO4 + 3 KOH → K3PO4 + 3 H2O

98 g 168 g 212 g 54 g



Step 5. Use known information to calculate reactant requirements:

1 mole: 98 g 56 g 212 g 18 g

H3PO4 + 3 KOH → K3PO4 + 3 H2O

98 g 168 g 212 g 54 g

For 20.0 g H3PO4:

98 g = 168 g X = 34.3 KOH 20 g X

For 30.0 g KOH:

98 g = 168 g X = 17.5 g H3PO4

X 30 g



Step 6. Analysis:

H3PO4 + 3 KOH → K3PO4 + 3 H2O

For 20.0 g H3PO4:

98 g = 168 g X = 34.3 KOH 20 g X

For 30.0 g KOH:

98 g = 168 g X = 17.5 g H3PO4

X 30 g

We need 34.3 g of KOH to completely react with 20.0 g of H3PO4. We only have 30.0 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.



Extension: Use the limiting reactant to solve for products:

H3PO4 + 3 KOH → K3PO4 + 3 H2O

98g 168 g 212 g 54 g

For K3PO4:

168 g = 212 g X = 37.9 g K3PO4

30 g X

For H2O :

168 g = 54 g X = 9.6 g H2O

30 g X



Extension: Use the limiting reactant to determine excess reactant.

H3PO4 + 3 KOH → K3PO4 + 3 H2O

98g 168 g 212 g 54 g

Since KOH is the Limiting Reactant, solve for required H3PO4:

168 g = 98 g X = 17.5 g H3PO4

30 g X

Start with: 20.0 g H3PO4

Need: 17.5 g Excess: 2.5 g H3PO4

stoichiometry the calculation of the quantities of chemical substances involved in chemical...

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pieces chocolate

pieces of chocolate

graham crackers

g of carbon

g of oxygen

g of product

limiting reactants definition

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