CHAPTER 03 Equations and Reaction Stoichiometry

CHAPTER THREE CHEMICAL

EQUATIONS & REACTION STOICHIOMETRY

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Chapter Three Goals1. 2. 3. 4. 5. 6. 7. 8. 9.

Chemical Equations Calculations Based on Chemical Equations The Limiting Reactant Concept Percent Yields from Chemical Reactions Sequential Reactions Concentrations of Solutions Dilution of solutions Using Solutions in Chemical Reactions Synthesis Question

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Chemical Equations

2. 3. 4.

Symbolic representation of a chemical reaction that shows: reactants on left side of reaction products on right side of equation relative amounts of each using stoichiometric coefficients

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Chemical Equations Attempt

to show on paper what is happening at the laboratory and molecular levels.

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Chemical Equations Look

at the information an equation provides:

Fe 2 O 3 + 3 CO 2 Fe + 3 CO 2

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Chemical Equations Look

at the information an equation provides:

Fe 2 O 3 + 3 CO 2 Fe + 3 CO 2 reactants yields products

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Chemical Equations Look

at the information an equation provides:

Fe 2 O 3 + 3 CO 2 Fe + 3 CO 2 reactants 1 formula unit 3 molecules yields 2 atoms products 3 molecules

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Chemical Equations Look

at the information an equation provides:

Fe 2 O 3 + 3 CO 2 Fe + 3 CO 2 reactants 1 formula unit 1 mole 3 molecules 3 moles yields 2 atoms 2 moles products 3 molecules 3 moles

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Chemical Equations Look

at the information an equation provides:

Fe 2 O 3 + 3 CO 2 Fe + 3 CO 2 reactants 1 formula unit 1 mole 159.7 g 3 molecules 3 moles 84.0 g yields 2 atoms 2 moles 111.7 g products 3 molecules 3 moles 132 g

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Chemical Equations

Law of Conservation of Matter

There is no detectable change in quantity of matter in an ordinary chemical reaction. Balanced chemical equations must always include the same number of each kind of atom on both sides of the equation. This law was determined by Antoine Lavoisier.

Propane,C3H8, burns in oxygen to give carbon dioxide and water. 3 8 2 2 2

C H +5O 3 CO + 4 H O

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Law of Conservation of Matter NH3

burns in oxygen to form NO & water You do it!

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Law of Conservation of Matter NH3

burns in oxygen to form NO & water5 2

2 NH 3 + O 2 2 NO + 3 H 2 O or correctly 4 NH 3 + 5 O 2 4 NO + 6 H 2 O12

Law of Conservation of Matter C7H16

burns in oxygen to form carbon dioxide and water. You do it!

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Law of Conservation of Matter C7H16

burns in oxygen to form carbon dioxide and water.

C 7 H16 + 11 O 2 7 CO 2 + 8 H 2 O

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Law of Conservation of Matter C7H16

burns in oxygen to form carbon dioxide and water.

C 7 H16 + 11 O 2 7 CO 2 + 8 H 2 O Balancing

equations is a skill acquired only with lots of practicework many problems

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Calculations Based on Chemical Equations Can

work in moles, formula units, etc. Frequently, we work in mass or weight (grams or kg or pounds or tons). Fe 2 O 3 + 3 CO 2 Fe + 3 CO 2

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Calculations Based on Chemical Equations Example

3-1: How many CO molecules are required to react with 25 formula units of Fe2O3?3 CO molecules 1 Fe 2 O 3 formula unit

? CO molecules = 25 formula units Fe 2 O 3 = 75 molecules of CO

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Calculations Based on Chemical Equations Example

3-2: How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?5

? Fe atoms = 2.50 10 formula units Fe 2 O 3

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Calculations Based on Chemical Equations Example

3-2: How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?5

? Fe atoms = 2.50 10 formula units Fe 2 O 3 2 Fe atoms = 1 formula units Fe 2 O 319

Calculations Based on Chemical Equations Example

3-2: How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?5

? Fe atoms = 2.50 10 formula units Fe 2 O 3 2 Fe atoms 5 = 5.00 10 Fe atoms 1 formula units Fe 2 O 320

Calculations Based on Chemical Equations Example

3-3: What mass of CO is required to react with 146 g of iron (III) oxide?

1 mol Fe 2 O 3 ? g CO = 146 g Fe 2 O 3 159.7 g Fe 2 O 3

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Calculations Based on Chemical Equations Example

3-3: What mass of CO is required to react with 146 g of iron (III) oxide?

1 mol Fe 2 O 3 3 mol CO ? g CO = 146 g Fe 2O 3 159.7 g Fe 2O 3 1 mol Fe 2O 3

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Calculations Based on Chemical Equations Example

3-3: What mass of CO is required to react with 146 g of iron (III) oxide?

1 mol Fe 2 O 3 3 mol CO ? g CO = 146 g Fe 2O 3 159.7 g Fe 2O 3 1 mol Fe 2O 3 28.0 g CO = 76.8 g CO 1 mol CO23

Calculations Based on Chemical Equations Example

3-4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?3 mol CO 2 ? g CO 2 = 0.540 mol Fe 2 O3 1 mol Fe 2 O3

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Calculations Based on Chemical Equations Example

3-4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?

3 mol CO 2 44.0 g CO 2 ? g CO 2 = 0.540 mol Fe 2 O 3 1 mol Fe 2 O 3 1 mol CO 2

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Calculations Based on Chemical Equations Example

3-4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide? 3 mol CO 2 44.0 g CO 2 ? g CO 2 = 0.540 mol Fe 2 O 3 1 mol Fe 2 O 3 1 mol CO 2 = 71.3 g CO 2

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Calculations Based on Chemical Equations Example

3-5: What mass of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams? You do it!

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Calculations Based on Chemical Equations Example

3-5: What mass of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams?

1 molCO2 1mol Fe2O3 ? g Fe2O3 = 8.65 g CO2 44.0 g CO2 3 mol CO2 159.7 g Fe 2O3 = 10.5 g Fe2O3 1 mol Fe2O328

Calculations Based on Chemical Equations Example

3-6: How many pounds of carbon monoxide would react with 125 pounds of iron (III) oxide? You do it!

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Calculations Based on Chemical Equations454 g Fe 2 O 3 ? lb CO = 125 lb Fe 2 O 3 1 lb Fe 2 O 3 1 mol Fe 2 O 3 3 mol CO 159.7 g Fe 2 O 3 1 mol Fe 2 O 3 28 g CO 1 lb CO = 65.7 lb CO 1 mol CO 454 g COYOU MUST BE PROFICIENT WITH THESE TYPES OF PROBLEMS!!! Now go to your text and work the problems assigned!

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Limiting Reactant Concept Kitchen

example of limiting reactant concept.

1 packet of muffin mix + 2 eggs + 1 cup of milk 12 muffins How

many muffins can we make with the following amounts of mix, eggs, and milk?

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Limiting Reactant Concept

Mix Packets Eggs Milk 1 1 dozen 1 gallon limiting reactant is the muffin mix 2 1 dozen 1 gallon 3 1 dozen 1 gallon 4 1 dozen 1 gallon 5 1 dozen 1 gallon 6 1 dozen 1 gallon 7 1 dozen 1 gallon limiting reactant is the dozen eggs

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Limiting Reactant Concept

Example 3-7: Suppose a box contains 87 bolts, 110 washers, and 99 nuts. How many sets, each consisting of one bolt, two washers, and one nut, can you construct from the contents of one box? 87 bolts 1 set = 87 sets 1 bolt 110 washers 1 set = 55 sets 2 washers 99 nuts 1 set = 99 sets 1 nut the maximum number we can make is 55 sets

( ( (

)

)

)

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determined by the smallest number

Limiting Reactant Concept Look

at a chemical limiting reactant situation. Zn + 2 HCl ZnCl2 + H2

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Limiting Reactant Concept

Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?

CS2 + 3 O 2 CO 2 + 2 SO 2

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Limiting Reactant Concept

Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?

CS2 + 3 O 2 CO 2 + 2 SO 2 1 mol36

3 mol

1 mol

2 mol

Limiting Reactant Concept

Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?

CS2 + 3 O 2 CO 2 + 2 SO 2 1 mol 3 mol 1 mol 2 mol 76.2 g 3(32.0 g) 44.0 g 2(64.1 g)37

Limiting Reactant Concept

Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?

CS2 + 3 O 2 CO 2 + 2 SO 2 1 mol CS2 ? mol SO 2 = 95.6 g CS2 76.2 g

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Limiting Reactant ConceptCS2 + 3 O 2 CO 2 + 2 SO 2 1 mol CS2 ? mol SO 2 = 95.6 g CS2 76.2 g 2 mol SO 2 64.1 g SO 2 =161 g SO 2 1 mol CS2 1 mol SO 2What do we do next? You do it!

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Limiting Reactant ConceptCS2 + 3 O 2 CO 2 + 2 SO 2 1 mol CS2 2 mol SO 2 64.1 g SO 2 ? mol SO 2 = 95.6 g CS2 = 161 g SO 2 76.2 g 1 mol CS2 1 mol SO 2 1 mol O 2 2 mol SO 2 64.1 g SO 2 ? mol SO 2 = 110 g O 2 = 147 g SO 2 32.0 g O 2 3 mol O 2 1 mol SO 2

Which is limiting reactant? Limiting reactant is O2. What is maximum mass of sulfur dioxide? Maximum mass is 147 g.

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Percent Yields from Reactions

Theoretical yield is calculated by assuming that the reaction goes to completion.

Determined from the limiting reactant