Chapter 4:

Chapter 4:

Balancing Equations and Stoichiometry

Key Terms & Concepts

Stoichiometry Chemical Equations

reactants and products balancing chemical equations

Chemical Calculations Limiting Reactant Theoretical and Percent Yield

Stoichiometry

Stoichiometry is the study of the quantitative nature of chemical formulas and chemical reactions.

Stoichiometry is one the the most essential tools in chemistry It allows to quantify everything from

global warming to drug manufacturing

Chemical Equations Chemical reactions are represented

in a concise manner by chemical equation

For example, when H2 burns in O2, H2O is formed. The chemical equation for this reaction

is:2 H2 + O2 2H2O

Chemical Equations

2 H2 + O2 2 H2O The compounds on the left of the

arrow are called “reactants” The compounds on the right of the

arrow are called “products” H2 and O2 are reactants, H2O is the

product

Chemical Equations

2 H2 + O2 2 H2O Notice that the number of atoms or each

element is equal on both sides of the equation 4 H, 2 0

All chemical equations must meet this requirement Chemical equations must be balanced!!

We balance equations by changing coefficients, not chemical formulas

Chemical Equations

Chemical Equations Consider the following chemical

equationCH4 + O2 CO2 + H2O

unbalanced Start with elements that only appear

in one compound on either side of the equation C and H are only in one compound on

either side C is balanced

Chemical Equations

CH4 + O2 CO2 + H2O

unbalanced 4 H’s in reactants, 2 H’s in products

Put coefficient of 2 in front of H2O 4 H’s in reactants, 4 H’s in products

CH4 + O2 CO2 + 2 H2O

unbalanced

Chemical Equations

CH4 + O2 CO2 + 2 H2O

unbalanced 2 O’s in reactants, 4 O’s in products

Put coefficient of 2 in front of O2

4 O’s in reactants, 4 O’s in products

CH4 + 2 O2 CO2 + 2 H2O

balanced

Chemical Equations

Chemical Equations

Consider this equation:C3H8 + O2 CO2 + H2O

3C + 8H + 2O 1C + 2H + 3O

Chemical Equations

Balance C and HC3H8 + O2 3CO2 + 4H2O

3C + 8H + 2O 3C + 8H + 10O

Chemical Equations

Balance OC3H8 + 5O2 3CO2 + 4H2O

3C + 8H + 10O 3C + 8H + 10O

Chemical Equations

Example 4.1Balance the following chemical equations(1) Mg + HCl MgCl2 + H2

(2) K + H2O KOH + H2

(3) CaCl2 + Na3PO4 Ca3(PO4)2 + NaCl

(4) NaN3 Na + N2

(5) C8H18 + O2 CO2 + H2O

Chemical Equations Example 4.1Balance the following chemical equations(1) Mg + 2 HCl MgCl2 + H2

(2) 2 K + 2 H2O 2 KOH + H2

(3) 3 CaCl2 + 2 Na3PO4 Ca3(PO4)2 + 6 NaCl

(4) 2 NaN3 2 Na + 3 N2

(5) 2 C8H18 + 25 O2 16 CO2 + 18 H2O

Chemical Equations Example 4.2

Write a balanced chemical equation for the following reactions

(1)ammonium nitrate decomposes to nitrogen gas, oxygen gas, and water

(2)iron reacts with oxygen gas and water to form iron(II) hydroxide

(3)ammonia reacts with oxygen gas to produce nitrogen monoxide and water

Chemical Equations

Example 4.2Write a balanced chemical equation for the following reactions

(1) 2 NH4NO3 2 N2 + O2 + 4 H2O

(2) 2 Fe + O2 + 2 H2O 2 Fe(OH)2

(3) 4NH3 + 5 O2 4 NO + 6 H2O

Chemical Calculations

2 H2 + O2 2 H2O

2 molecules 1 molecule 2 molecules

2(6.022x1023) molecules 6.022x1023 molecules 2(6.022x1023) molecules

2 mol 1 mol 2 mol

Stoichiometric coefficients can be interpreted as either number of molecules or number of moles.


Example 4.3How many moles of water can be produced from 5.25 mol O2?

OH mol 10.5O mol 1

OH mol 2O mol 5.25 2

2

22


Example 4.4How many moles of oxygen are required to completely react with 8.50 moles of butane, C4H10?

OH 10CO 8O 13HC 2 222104

2104

2104 O mol 55.3

HC mol 2

O mol 13HC mol 8.50


We can’t directly measure moles. We can measure mass.

We can use the stoichiometric coefficients of a reaction to determine the mass relationships.

However, we must always convert mass to moles. We cannot directly compare the masses of

reactants and products. We can only compare the moles of

reactants and products.


The general scheme is:


Example 4.5Geranyl formate is used as a synthetic rose essence in cosmetics. The compound is prepared from formic acid and geraniol:

HCOOH + C10H18O C11H18O2 + H2O

A chemist needs to make some geranyl formate for a batch of perfume. How many grams of geranyl formate can a chemist make from 375g of geraniol?


g 443mol 1

g 182.3

OHC mol 1

OHC mol 1

g 154.2

mol 1g 375

g/mol 182.3OHC

g/mol 154.2 OHC

1810

21811

21811

1810


Example 4.6Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide. The lithium hydroxide reacts with gaseous carbon dioxide to form solid lithium carbonate and liquid water. How many grams of carbon dioxide can be absorbed by each 1.00 g of lithium hydroxide?2 LiOH (s) + CO2 (g) Li2CO3 (g) + H2O

(l)


Example 4.62 LiOH (s) + CO2 (g) Li2CO3 (g) + H2O (l)

22

2 CO g 0.919CO mol 1

g 44.01

LiOH mol 2

CO mol 1

g 23.95

LiOH mol 1LiOH g 1.00

Theoretical and Percent Yield

The amount of product that can be produced from a given amount of reactants is the theoretical yield.

However, no reaction goes to actual completion. The amount of products that is actually produced from a given amount of reactants is the actual yield.•Some reactants may not react•Reactants may react in an undesired way

(side reactions) •May be difficult to remove products from

pot


The extent of the desired reaction is typically reported as the percent yield.

100%Yield lTheoretica

Yield Actual YieldPercent


Example 4.7Look back at Example 4.5. If the chemist starts with 375g of geraniol and collects 417g of purified product, what is the percent yield of the synthesis?

%1.94%100g 443

g 417


Example 4.825.0 g of sodium metal is burned in an excess of chlorine gas. What is the theoretical yield of sodium chloride? If 54.8 g of sodium chloride is actually produced, what is the percent yield of the reaction?

2 Na + Cl2 2 NaCl


yield al theoretic theis This

NaCl g 63.5NaCl mol 1

g 58.4

Na mol 2

NaCl mol 2

g 23.0

Na mol 1Na g 25.0


yieldpercent theis This

86.3% 100%NaCl g 63.5

NaCl g 54.8


Example 4.9Titanium is a strong, lightweight, corrosion-resistant metal that is used in aeronautics and bicycle frames. It is prepared by the reaction of titanium (IV) chloride with molten magnesium between 950C and 1150C.TiCl4(g) + 2 Mg(l) Ti(s) + 2 MgCl2(l)

In a certain industrial process 3.54x107 g of TiCl4 are reacted with 1.13x107 g of Mg. (a)Calculate the theoretical yield of Ti in grams. (b)Calculate the percent yield if 7.91x106 g of Ti are actually produced.


yield al theoretic theis Ti g 108.94 and

reactant limiting theis TiCl

Ti g 101.11Ti mol 1

g 47.9

Mg mol 2

Ti mol 1

g 24.3

Mg mol 1Mg g 101.13

Ti g 108.94Ti mol 1

g 47.9

TiCl mol 1

Ti mol 1

g 189.7

TiCl mol 1TiCl g 103.54

6

4

7

7

6

4

44

7


yieldpercent theis This

88.5%100%Ti g 108.94

Ti g 107.916

6

Limiting Reactants

Most reactions do not occur with stoichiometric equivalent amounts of each reactant.•One reactant is used up first•This reactant is the limiting reactant

because it limits the amount of products that can be formed

Limiting Reactants

Consider the “ham sandwich” example•one sandwich is made from one slice of

ham, one slice of cheese and two slices of bread

•How many ham sandwiches can be made from six slices of ham, seven slices of cheese and 14 slices of bread?

•What is the limiting reactant?

Limiting Reactants

Limiting Reactants

If a problem gives specific amounts of two or more reactants it is a limiting reactant problem.

Determine the amount of product that can be formed from each reactant•The reactant which produces the smallest

amount of product is the limiting reactant•The remaining reactants are said to be in

excess

Limiting Reactants

Example 4.10How many moles of water can be formed when 10.0 moles of H2 reacts with 4.50 moles of O2? What is the limiting reactant?

Limiting Reactants

Example 4.10

2 H2 + O2 2 H2O

formed becan OH of mol 9.00 and

reactant limiting theis O

OH mol 9.00O mol 1

OH mol 2O mol 4.50

OH mol 10.0H mol 2

OH mol 2H mol 10.0

2

2

22

22

22

22

Limiting Reactants

Example 4.11Solutions of sulfuric acid and lead (II) acetate react to form solid lead (II) sulfate and aqueous acetic acid. If 15.0 g of sulfuric acid and 15.0 g of lead (II) acetate are mixed, calculate the number of grams of lead (II) sulfate that can be produced. Also calculate the number of grams of the excess reagent remaining after the reaction is completed.

Limiting Reactants

formed becan PbSO of g 14.0 and

reactant limiting theis COO)Pb(CH

PbSO g 0.41PbSO mol 1

g 303.3

COO)Pb(CH mol 1

PbSO mol 1

g 25.33

COO)Pb(CH mol 1COO)Pb(CH g 15.0

PbSO g 46.4PbSO mol 1

g 303.3

SOH mol 1

PbSO mol 1

g 1.98

SOH mol 1SOH g 15.0

4

23

4423

4

2323

4442

4

4242

H2SO4 + Pb(CH3COO)2 PbSO4 + 2 CH3COOH

Limiting Reactants

excess.in SOH of g 10.5 g 4.52 - g 15.0

leavingreaction in the used are SOH of g 4.52

SOH g 52.4SOH mol 1

g 98.1

COO)Pb(CH mol 1

SOH mol 1

g 25.33

COO)Pb(CH mol 1COO)Pb(CH g 15.0

42

42

424223

42

2323

End-of-Chapter Exercises

Suggested End-of-Chapter Exercises5, 9, 10, 14, 17, 20, 22, 28

Chapter 4:

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