Chapter 3: Stoichiometry

Chapter 3: StoichiometryChapter 3: Stoichiometry

● “measuring elements”

● Must account for ALL atoms in a chemical reaction

+ →

H2

+ O2

→ H2O22


CO + O2

→ CO2

22

+ →

CH4

+ Cl2

→ CCl4

+ HCl

→+ +

4 4


C2H

4+ O

2→ CO

2+ H

2O2 23

2 Al + 6 HCl → 2 AlCl3

+ 3 H2

2 NH4NO

3→ 2 N

2 + O

2 + 4 H

2O


Three basic reaction types:

● Combination Reactions

● Decomposition Reactions

● Combustions (in air)


Combination Reactions

Two or more reactants combine to form a single product

C (s) + O2 (g) → CO

2 (g)

Decomposition Reactions

A single reactant breaks into two or more products

2 KClO3 (s) → 2 KCl (s) + 3 O

2 (g)


Combustions in Air = reactions with oxygen

Write the balanced reaction equation for the combustion of magnesium tomagnesium oxide:


Combustions of Hydrocarbons in Air

= reactions with oxygen to form carbon dioxide and water(complete combustion)

Write the balanced reaction equation for the combustion of C2H

4 gas

C2H

4 (g)


C2H

4 (g) + O

2 (g) → CO

2 (g) + H

2O (g)2 23

+ → +

How many C2H

4 molecules are in the flask?

● If you know the weight of one molecule of C2H

4

and the total weight of gas in the flask, you cancalculate the number of molecules in the flask


Molecular weight / Formula weight:

=> sum of all atomic weights in molecular formula

MW of C2H

4

FW of Mg(OH)2

for molecular compounds

for ionic compounds


Ca(NO3)

2

Type of compound:

Ions:

Total number of oxygen atoms:

Name:


Ca(NO3)

2

Percentage of oxygen, by mass:

(1) total mass of Ca(NO3 )

2 in amu

(2) mass of oxygen in compound, in amu

(3) percentage of oxygen


Number of individual molecules are difficult to deal with

=> definition of a “package” of molecules or particles

1 dozen eggs = 12 individual eggs

1 soda sixpack = 6 individual bottles of soda

1 mole of molecules = 6.02 x 1023 individual molecules

Avogadro's Number.

... .... ..

. ...... . ..

..

. . ...

...

.....

...



How many moles of eggs are in an egg carton that holds a dozen eggs?

12 individual eggs ×



How many moles of eggs are in an egg carton that holds a dozen eggs?


Ca(NO3)

2

How many moles of oxygen are in 2.4 moles of Ca(NO3)

2 ?


Molar Mass

= mass of one mole of a substance in grams

FW or MW of substance in amu's = mass of 1mole of substance in grams

FW of Ca(NO3)

2 = 164.1 amu

Molar Mass of Ca(NO3)

2 = 164.1 g/mol

MW of O2 = 2 x 16.0 amu = 32 amu

Molar Mass of O2 = 32 g/mol


What is the mass in grams of 0.527 moles of CH3OH?

(1) determine molar mass of CH3OH

32 g/mol

(2) use MM to convert moles into grams

g CH3OH16.9


mol → gram←

Molar Mass

How many hydrogen atoms are in 4.5 g of CH3OH?

= mol CH3OH

= mol H atoms

= H atoms

0.14

0.56

3.4 x 1023


Quantitative Information from Balanced Equations

CO + O2

→ CO2

22

How many grams of CO2 would be produced by the combustionof 2 moles of CO?

+ →

2 moles CO + 1 mole O2 → 2 moles CO2

2 x 28 g + 32 g → 2 x 44 g

= 88 g


Quantitative Information from Balanced Equations

CO + O2

→ CO2

22

You can write a series of stoichiometric factors for this reaction:

2mol CO

1 mol O2 2mol CO

1 mol O2 1 mol O2

2 mol CO2


C2H

4 (g) + O

2 (g) → CO

2 (g) + H

2O (g)2 23

How many grams of H2O are formed from the complete combustionof 2 g of C2H4?

28 g/mol 18 g/mol

1 mol 2 mol

= 2.6 g H2O


Summary

1) determine equation for the reaction

2) balance equation

4) determine MW/FW of substances involved

5) determine stoichiometric factors from balanced equation

3) formulate problem:how much of A => gets converted into how much of B


Limiting Reactants

+

+

Limiting Reactant

2 of these willbe left over

“in excess”


Limiting Reactants

Limiting Reactant

- limits the amount of product that can be formed

- reacts completely (disappears during the reaction)

- other reactants will be left over, i.e. in excess


Limiting Reactants

N2 + 3 H

2 → 2 NH

3

3 mol N2 , 6 mol H

2Available:

How much H2 would we need to completely react 3 mol N2:

= 9 mol H2

How much NH3 can we form with the available reagents?

= 4 mol NH3


Limiting Reactants

N2 + 3 H

2 → 2 NH

3

3 mol N2 , 6 mol H

2Available:

How much N2 is left over (in excess)?

= 1 mol N2


Limiting Reactants

Available:

= 0.75 mol Cl2

How much AlCl3 can we form with the available reagents?

= 0.5 mol AlCl3

2 Al + 3 Cl2 → 2 AlCl

3

0.5 mol Al , 2.5 mol Cl2

How much Cl2 would we need to completely react 0.5 mol Al:


Theoretical Yield

Available:

What mass do 0.5 mol AlCl3 correspond to?

= 67 g AlCl3

2 Al + 3 Cl2 → 2 AlCl

3


The maximum mass of product that can be formed is thetheoretical yield


Theoretical Yield

Available:

Fritz does the reaction with the available reagents he only ends up with 34g. What is the % yield of the reaction?

2 Al + 3 Cl2 → 2 AlCl

3


% yield = actual yieldtheoretical yield

× 100 %

%1006734

% gg

yield = 51 %


Summary

Determine availabequantity of reactants

in moles

Determine if one ofthe reactants is a limiting reactant

Determine the maximum

# of moles of productthat can be formed

Convert into grams ofproduct

(theoreticalyield)

Compare with actual

amount of productrecovered

(actual yield)

Determine % yieldof the reaction


2 CH3OH + 3 O2 → 2 CO2 + 4 H2O

Consider the combustion of methanol:

What is the theoretical yield of water if 44 g of methanol are reacted with 128 g of oxygen?

OHCHmol 34.1

20.4 Omol


2 CH3OH + 3 O2 → 2 CO2 + 4 H2O

= 2.1 mol O2

available(initial) moles 1.4 mol 4.0 mol O2


2 CH3OH + 3 O2 → 2 CO2 + 4 H2O

(3) how many product moles can be formed with limiting reactant ?

= 1.4 mol CO2

= 2.8 mol H2O


2 CH3OH + 3 O2 → 2 CO2 + 4 H2O

(4) What is the mass of H2O formed (theoretical yield )?

OH 50g 2

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Chapter 3: Stoichiometry