unit 6 stoichiometry chapter 9. chapter 9 stoichiometry

Unit 6

StoichiometryChapter 9Chapter 9

CHAPTER 9Stoichiometry

What is Stoichiometry?

• Stoichiometry is the branch of chemistry that deals withthe mass relationships of elements in a chemical reaction.

• Stoichiometry calculationsalways start with a balanced chemical equation.

Chapter 9 – Section 1: Introduction to Stoichiometry

Mole Ratio

• A mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reactionExample: 2Al2O3(l) → 4Al(s) + 3O2(g)

Mole Ratios:

2 mol Al2O3 2 mol Al2O3 4 mol Al

4 mol Al 3 mol O2 3 mol O2


Molar Mass (Review)• Molar Mass is the mass of one mole of a pure

substance.• Molar mass units

are g/mol.• The molar mass of

an element is the same number as its atomic mass, only the units are different.

• Example: Molar mass of water (H2O).• (2 mol H x 1.01) + (1 mol O x 16.00) = 18.02 g/mol.


Mole to Mole Conversions• Using the mole ratio, you can convert from moles

of one substance to moles of any other substance in a chemical reaction.

Example:• For the reaction N2 + 3H2 → 2NH3, how many moles

of H2 are required to produce 12 moles of NH3?

Chapter 9 – Section 2: Ideal Stoichiometric Calculations

This is whatwe’re given:

Use mole ratio as a conversion factor

12 mol NH33 mol H2

2 mol NH3

18 mol H2=x

Mole/Mole ConversionsSample Problem 1

In a spacecraft, the CO2 exhaled by astronauts can be removed by the following reaction with LiOH:

CO2(g) + 2LiOH(s) → Li2CO3(s) + H2O(l)How many moles of lithium hydroxide are required to react with 20 mol CO2, the average amount exhaled by a person each day?

Solution:Given Conversion factor

20 mol CO22 mol LiOH1 mol CO2

40 mol LiOH=x


Mole to Mass Conversions• To convert from moles of one substance to grams

of another, you need 2 conversion factors: 1. mole ratio.2. molar mass of the unknown.

• To set up your conversion factor, always put the units you have on the bottom and the units you need on the top.


Example:Given the equation: 2Mg(s) + O2(g)→ 2MgO(s),

Calculate the mass in grams of magnesium oxide which is produced from 2.00 mol of magnesium.

mol MgO

mol MgO

Mole to Mass Conversions (continued)


GivenMole Ratio

2.00 mol Mgmol Mg

80.6 g MgO=x x

2nd C.F.Molar Mass

of the Unknown

1st C.F.

2

2

1

g MgO40.3

Mass to Mole Conversions• To convert from grams of one substance to moles of

another, you need 2 conversion factors: 1. molar mass of the given.2. mole ratio.

• Mass to mole conversion factors are the inverse of mole to mass conversion factors.


Example:Given the equation: 2HgO(s) → 2Hg(s) + O2(g),

How many moles of HgO are needed to produce 125g of O2?

mol O2

mol O2

Mass to Mole Conversions (continued)


Given

Molar Massof the Given

125 g O2

g O2

7.81 mol HgO=x x

2nd C.F.Mole Ratio

1st C.F.

32

1

1

mol HgO2

Mole/Mass ConversionsSample Problem 1

The balanced equation for photosynthesis is as follows:

6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g)What mass, in grams, of glucose (C6H12O6) is produced when 3.00 mol of water react with carbon dioxide?

Solution:


mol C6H12O6

mol C6H12O6

GivenMole Ratio

3.00 mol H2Omol H2O

90.1 g C6H12O6

=x x

2nd C.F.Molar Mass

of the Unknown

1st C.F.

6

1

1

g C6H12O6180.2

Mole/Mass ConversionsSample Problem 2

The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

If the reaction is run using 824 g NH3 and excess oxygen, how many moles of NO are formed?

Solution:


mol NH3

mol NH3

Given


824 g NH3

g NH3

48.5 mol NO=x x

2nd C.F.Mole Ratio

1st C.F.

17.0

1

4

mol NO4

Mass to Mass Conversions• To convert from grams of one substance to

grams of another, you need 3 conversion factors:

1. molar mass of the given.2. mole ratio.3. molar mass of the unknown.


Example:Given the equation: 2HgO(s) → 2Hg(s) + O2(g),

How many grams of HgO are needed to produce 45g of O2?

mol O2

mol O2

Mass to Mass Conversions (continued)


Given


45 g O2

g O2

609.2 g HgO

=x x

2nd C.F.Mole Ratio

1st C.F.

32

1

1

mol HgO2

mol HgOx

3rd C.F.Molar Mass ofthe Unknown

1

g HgO216.6

Mass/Mass ConversionsSample Problem 1

Tin(II) fluoride, SnF2, is used in some toothpastes. It is made by the following reaction :

Sn(s) + 2HF(g) → SnF2(s) + H2(g)

How many grams of SnF2 are produced from the reaction of 30.0 g HF with Sn?

Solution:


mol HF

mol HF

GivenMolar Massof the Given

30.0 g HFg HF

=x x

2nd C.F.Mole Ratio

1st C.F.

20.0

1

2

mol SnF21

mol SnF2

x

3rd C.F.Molar Mass ofthe Unknown

1

g SnF2156.7 117.5 g SnF2

Limiting Reactants

• When combining 2 or more different things to make a product, you have to stop when one of the things is used up.

• For example, no matter how many tires there are, if there are only 8 car bodies, then only 8 cars can be made.

Chapter 9 – Section 3: Limiting Reactants and Percentage Yield

Limiting Reactants (continued)

• The limiting reactant is the reactant that limits the amount of product formed.

• The excess reactant is the substance that is not used up completely.

• Once the limiting reactant is used up, a chemical reaction will stop.


Limiting Reactants (continued)

Example:Silicon dioxide reacts with hydrogen fluoride according

to the following equation:SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l)

If 6.0 mol HF is added to 4.5 mol SiO2, which is the limiting reactant?


Set up 2 equations, one for eachgiven:

Mole ratio C.F.

4.5 mol SiO21 mol SiF4

1 mol SiO2

4.5 mol SiF4=x

6 mol HF 1 mol SiF4

4 mol HF1.5 mol SiF4=x 1.5 mol SiF46 mol HF

Limiting Reactant is HF

The limitingreactant

makes the least product.

Limiting ReactantsSample Problem 1

According to the following reaction :CS2(l) + 3O2(g) → CO2(g) + 2SO2(g)

a.If 1 mol CS2 reacts with 1 mol O2, identify the limiting reactant.

b.How many moles of excess reactant remain?


1 mol O2 2 mol SO2

3 mol O2

0.67 mol SO2=x

1 mol CS22 mol SO2

1 mol CS2

2 mol SO2=

1 mol O2

x

0.67 mol SO2

Limiting Reactant is O2

0.67 mol SO2 1 mol CS2

2 mol SO2

0.33 mol CS2=xWork backwards from actual mol of product formed

Mol of excessreactant thatwas used up

1 mol CS2 - 0.33 mol CS20.67 mol CS2=

Molesremaining

Percentage Yield

• The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant (found with stoichiometry).

• The actual yield of a product is the measured amount of that product obtained from a reaction (given in problem).

• The percentage yield is the ratio of the actual yield to the theoretical yield, multiplied by 100.


Now use the formula

mol C6H5Cl

Percentage Yield (continued)

Example:Given the following equation:

C6H6 (l) + Cl2(g) → C6H5Cl(l) + HCl(g)When 36.8 g C6H6 react with excess Cl2, the actual yield of

C6H5Cl is 38.8 g. What is the percentage yield of C6H5Cl?


mol C6H6

mol C6H636.8 g C6H6g C6H6

=x x78.01

11

mol C6H5Clx

1g C6H5Cl112.5 53.1 g

C6H5Cl

Set up a mass to mass conversionThis is the

Theoretical Yield

Percent Yield

=Actual

Theoreticalx 100 =

38.8 g C6H5Cl53.1 g C6H5Cl

x 100 = 73.1%

Percentage YieldSample Problem 1

According to the following reaction :CO(g) + 2H2(g) → CH3OH(l)

If the typical yield is 80%, what mass of CH3OH should be expected if 75.0 g of CO reacts with excess hydrogen gas?


Multiply theoreticalyield by percentageto get actual yield

mol CH3OHmol CO

mol CO75.0 g COg CO

=x x28.01

11

mol CH3OHx

1g CH3OH32.0 85.7 g

CH3OH

Set up a mass to mass conversionThis is the

Theoretical Yield

85.7 g CH3OH x 80% = 68.6 g CH3OH

unit 6 stoichiometry chapter 9. chapter 9 stoichiometry

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