chapter 4 chemical reactions. 4.1 chemical reactions and chemical equations

Chapter 4

Chemical Reactions

4.1 Chemical Reactions and Chemical Equations

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Chemical Equations

Concise representations of chemical reactions

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Anatomy of a Chemical Equation

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

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Anatomy of a Chemical Equation

Reactants appear on the left side of the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

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Anatomy of a Chemical Equation

Products appear on the right side of the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

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Anatomy of a Chemical Equation

The states of the reactants and products are written in parentheses to the right of each compound.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

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Anatomy of a Chemical Equation

Coefficients are inserted to balance the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

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Subscripts and Coefficients Give Different Information

• Subscripts tell the number of atoms of each element in a molecule

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Subscripts and Coefficients Give Different Information

• Subscripts tell the number of atoms of each element in a molecule

• Coefficients tell the number of molecules

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Stoichiometric Calculations

The coefficients in the balanced equation give the ratio of moles of reactants and products

Hints for Balancing

• If an element occurs in only one compound on each side of the equation, balance that first

• Balance free elements last• Balance polyatomic ions as a group unit if they

remain unchanged• If you get stuck, try doubling everything!

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 4 Slide 13 of 24

Balancing Equations

• Never introduce extraneous atoms to balance.

NO + O2 → NO2 + O

Never change a formula for the purpose of balancing an equation.

NO + O2 → NO3

An equation can be balanced only by adjusting the coefficients of formulas.

Example 4-1 A

• Balance the equations:H3PO4 + CaO Ca3(PO4)2 + H2O

C3H8 + O2 CO2 + H2O

Example 4-2A

• Write a balanced equation to represent the reaction of mercury(II) sulfide and calcium oxide to produce calcium sulfide, calcium sulfate and mercury metal.

(s) Solid(l) Liquid(g) Gas(aq) aqueousOver the arrow: catalyst, heat or conditions needed

4.2 Chemical Equations and Stoichiometry

To Measure Elements!

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Stoichiometric Calculations

From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)

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Stoichiometric Calculations

Starting with 1.00 g of C6H12O6…

we calculate the moles of C6H12O6…

use the coefficients to find the moles of H2O…

and then turn the moles of water to grams

C6H12O6 + 6 O2 6 CO2 + 6 H2O

Switching Substances

YOU CAN ONLY SWITCH SUBSTANCES IN MOLES!!

Must use balanced equation!

Example 4-3A

• How many moles of O2 are produced from the decomposition of 1.76 moles of potassium chlorate?2 KClO3(s) 2 KCl(s) + 3O2(g)

Example 4-4A

How many grams of magnesium nitride are produced by the reaction of 3.82 grams of Mg and excess N2?

Example 4-5A

For the reactionNH3 + O2 N2 + H2O

How many grams of NH3 are consumed per gram of O2?

4.3 Chemical Reactions in Solution

Molarity

Solvent: does the dissolvingSolute: is dissolvedSolution: Solute and SolventMolarity = Amount of Solute (moles)

Volume of Solution (Liters)

Example 4-8A

A 22.3 gram sample of acetone is dissolved in enough water to produced 1.25L of solution. What is the molarity of the acetone in this solution?

Example 4-9A

An aqueous solution saturated with NaNO3 at 25°C is 10.8 M NaNO3. What mass of NANO3 is present 125mL of this solution at 25°C?

Dilutions from Stock

• M1V1 = M2V2

Solution Dilution

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 4 Slide 29 of 24

Figure 4-6

•Visualizing the dilution of a solution

Mi × Vi = ni

Mi Vi Mf Vf

= nf = Mf × Vf

M = n

V

Mi × ViMf = Vf

= Mi

Vi

Vf

Example 4-10A

A 15.00mL sample of 0.450M K2CrO4 is diluted to 100.00mL. What is the concentration of the new solution?

Example 4-11A

How many milliliters of 0.250 M K2CrO4 must be added to excess AgNO3(aq) to produce 1.50 g Ag2CrO4?

4.4 Determining the Limiting Reactant

Limiting Reactants

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How Many Cookies Can I Make?

• You can make cookies until you run out of one of the ingredients

• Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat)

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How Many Cookies Can I Make?

• In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make

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Limiting Reactants

The limiting reactant is the reactant present in the smallest stoichiometric amount

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Limiting Reactants

• The limiting reactant is the reactant present in the smallest stoichiometric amount– In other words, it’s the reactant you’ll run out of first (in

this case, the H2)

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Limiting Reactants

In the example below, the O2 would be the excess reagent

Example 4-12A

• If 215 g P4 is allowed to react with 725 g Cl2 how many grams of PCl3 are formed?

P4 + 6Cl2 4PCl3

Example 13A

• From 12A, which reactant is in excess and what mass of the reactant remains after the reaction to produce PCl3?

4.5 Other Practical Matters in Reaction Stoichiometry

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Theoretical Yield

• The theoretical yield is the amount of product that can be made– In other words it’s the amount of product possible

as calculated through the stoichiometry problem

• This is different from the actual yield, the amount one actually produces and measures

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Percent Yield

A comparison of the amount actually obtained to the amount it was possible to make

Actual YieldTheoretical YieldPercent Yield = x 100

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Example1.20 g Al reacts with 2.40 g I2. What is the limiting reactant?

1st find the reactant that produces the smallest amount of product.

2Al + 3 I2 2AlI3

1.20 g Al x 1mol Al x 2 mol AlI3 x 407.7 g AlI3 = 18.1 g AlI3

27 g Al 2 mol Al 1 mol AlI3

2.40 g I2 x 1 Mol I2 x 2 mol AlI3 x 407.7 g AlI3 = 2.57 g AlI3

253.8 g I2 3 mol I2 1mol AlI3

I2 is limiting and 2.57 g is the theoretical yield

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Theoretical Yield

% Yield= actual Yield/ theoretical yield x 100

Say the actual yield was 2.05 g

Then

%8.7910057.2

05.2