STOICHIOMETRY
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STOICHIOMETRY Calculations Based on Chemical Equations
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STOICHIOMETRY. Calculations Based on Chemical Equations. Iron (III) oxide reacts with carbon monoxide to form iron and carbon dioxide. 3. 2. 3. Fe 2 O 3 +CO Fe+CO 2. How many CO molecules are required to react with 25 particles of Fe 2 O 3 ?. - PowerPoint PPT Presentation
Transcript of STOICHIOMETRY
STOICHIOMETRYCalculations Based on
Chemical Equations
Iron (III) oxide reacts with carbon monoxide to form iron and carbon dioxide.Fe2O3 + CO Fe + CO2
32
32
OFeunit form. 1
CO molecules 3
1
OFe units form. 25
3 2 3
How many CO molecules are required to react with 25 particles of Fe2O3?
CO molecules 75
Iron (III) Oxide reacts with carbon monoxide to form iron and carbon dioxide.Fe2O3 + CO Fe + CO2
32
325
OFeunit form. 1
Fe atoms 2
1
OFe units form.10 2.5
3 2 3
How many iron atoms can be produced by the reaction of 2.5 x 105 particles of Fe2O3?
Fe atoms 10 .05 5
STOICHIOMETRIC RELATIONSHIPS
• Formulas can also represent MOLES of substances involved in chemical reactions.
• Equations define reaction ratios, i.e. the molar ratios of reactants and products
What mass of CO is required to react with 146 grams of iron (III) oxide?
32
3232
OFe g 6.159
OFe mol 1OFe g 146 32OFe mol 1
CO mol 3
CO mol 1
CO g 28
Fe2O3 + 3CO 2Fe + 3CO2
CO g 76.8
What mass of iron (III) oxide is required to produce 8.65 grams of carbon dioxide?
2
22
CO g 44
CO mol 1CO g .658 2
32
CO mol 3
OFe mol 1 32
32
OFe mol 1
OFe g 159.6
Fe2O3 + 3CO 2Fe + 3CO2
32OFe g 10.5
Limiting Reactants (Reagents)
and Percent Yield
Calculations need to be based on the limiting reactant.• Example 1: Suppose a box contains
87 bolts, 110 washers and 99 nails. How many sets of 1 bolt, 2 washers and 1 nail can you use to create? What is the limiting factor?
55 sets; washers limit the amount
Calculations need to be based on the limiting reactant.
• Example 2: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g carbon disulfide with 100. g oxygen?CS2 + O2 CO2 + SO2
Start by balancing the equation…
Calculations need to be based on the limiting reactant.• Example 2: What is the maximum
mass of sulfur dioxide that can be produced by the reaction of 95.6 g carbon disulfide with 100. g oxygen?CS2 + 3O2 CO2 + 2SO2
Now solve the problem…
• Example 2: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g carbon disulfide with 100. g oxygen?CS2 + 3O2 CO2 + 2SO2
2
2
2
2
2
2
22 1611
1.64
1
2
2.76
16.95SOg
SOmol
SOg
CSmol
SOmol
CSg
CSmolCSg
22
2
2
2
2
22 1341
1.64
3
2
0.32
1.100SOg
SOmol
SOg
Omol
SOmol
Og
OmolOg
Which reactant is LIMITING?
• Example 2: What is the mass of sulfur dioxide that can be produced by the reaction of 95.6 g carbon disulfide with 100. g oxygen?CS2 + 3O2 CO2 + 2SO2
2
2
2
2
2
2
22 1611
1.64
1
2
2.76
16.95SOg
SOmol
SOg
CSmol
SOmol
CSg
CSmolCSg
22
2
2
2
2
22 1341
1.64
3
2
0.32
1.100SOg
SOmol
SOg
Omol
SOmol
Og
OmolOg
O2 limits the amount of SO2 that can be produced. CS2 is in excess.
• Example 2: What is the mass of sulfur dioxide that can be produced by the reaction of 95.6 g carbon disulfide with 100. g oxygen?CS2 + 3O2 CO2 + 2SO2
2
2
2
2
2
2
22 1611
1.64
1
2
2.76
16.95SOg
SOmol
SOg
CSmol
SOmol
CSg
CSmolCSg
22
2
2
2
2
22 1341
1.64
3
2
0.32
1.100SOg
SOmol
SOg
Omol
SOmol
Og
OmolOg
134 g of SO2 can be produced in this reaction.
Calculations need to be based on the limiting reactant.• Example 3: What mass of CO2
could be formed by the reaction of 8.0 g CH4 with 48 g O2?
CH4 + O2 CO2 + H2O
Start by balancing the equation…
Calculations need to be based on the limiting reactant.
• Example 3: What mass of CO2 could be formed by the reaction of 8.0 g CH4 with 48 g O2?
CH4 + 2O2 CO2 + 2H2O
Now solve the problem…
• Example 3: What mass of CO2 could be formed by the reaction of 8.0 g CH4 with 48 g O2?
CH4 + 2O2 CO2 + 2H2O
2
2
2
4
2
4
44 221
0.44
1
1
0.16
10.8COg
COmol
COg
CHmol
COmol
CHg
CHmolCHg
22
2
2
2
2
22 331
0.44
2
1
0.32
148COg
COmol
COg
Omol
COmol
Og
OmolOg
Which reactant is LIMITING?
• Example 3: What mass of CO2 could be formed by the reaction of 8.0 g CH4 with 48 g O2?
CH4 + 2O2 CO2 + 2H2O
CH4 limits the amount of CO2 that can be produced. O2 is in excess.
22
2
4
2
4
44 221
0.44
1
1
0.16
10.8COg
COmol
COg
CHmol
COmol
CHg
CHmolCHg
22
2
2
2
2
22 331
0.44
2
1
0.32
148COg
COmol
COg
Omol
COmol
Og
OmolOg
Many chemical reactions do not go to completion (reactants are not completely converted to products).
Percent Yield: indicates what percentage of a desired product is obtained.
100yieldltheoreticayieldalexperiment
Yield%
• So far, the masses we have calculated from chemical equations were based on the assumption that each reaction occurred 100%.
• The THEORETICAL YIELD is the yield calculated from the balance equation.
• The ACTUAL YIELD is the amount “actually” obtained in an experiment.
• Look back at Example 2. We found that 134 g of SO2 could be formed from the reactants.
• In an experiment, you formed 130 g of SO2. What is your percent yield?
%97100134
130%
g
gYield
Example: A 10.0 g sample of ethanol, C2H5OH, was boiled with excess acetic acid, CH3COOH, to produce 14.8 g of ethyl acetate, CH3COOC2H5. What percent yield of ethyl acetate is this?
CH3COOH + C2H5OH CH3COOC2H5 + H2O
5H
2COOC
3CH mol 1
5H
2COOC
3CH g 88
OH5
H2
C mol 1
5H
2COOC
3CH mol 1
OH5
H2
C 46g
OH5
H2
C mol 1OH5
H2
C 10.0g
5H
2COOC
3CH g1.19
Example: A 10.0 g sample of ethanol, C2H5OH, was boiled with excess acetic acid, CH3COOH, to produce 14.8 g of ethyl acetate, CH3COOC2H5. What percent yield of ethyl acetate is this?
100yieldltheoretica
yieldalexperimentYield%
100g 19.1
g 14.8Yield%
%5.77Yield%
