Chapter Outline - Resources
Transcript of Chapter Outline - Resources
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Chapter 7: Stoichiometry - Mass
Relations in Chemical Reactions
How do we balance chemical equations?
How can we used balanced chemical equations to
relate the quantities of substances consumed and
produced in chemical reactions?
How can we determine a compound’s elemental
composition and chemical formula?
→ →
Chapter Outline
7.1 Chemical reactions and the Conservation of Mass
7.2 Balancing Chemical Equations
7.3 Combustion Reactions
7.4 Stoichiometric Calculations and the Carbon Cyle
7.5 Percent Composition and Empirical Formulas
7.6 Empirical and Molecular Formulas Compared
7.7 Combustion Analysis
7.8 Limiting Reactants and Percent Yield
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Law of Conservation of Mass
The law of conservation of mass states that the sum
of the masses of the reactants of a chemical equation
is equal to the sum of the masses of the products.
C(s) + O2(g) → CO2(g)
12 g 32 g 44 g
1 mole 1 mole 1 mole
+ =
2 C(s) + O2(g) → 2 CO(g)
2 x 12 = 24 g
2 mole 1 mole 2 mole
32 g + 2 x 28 = 56 g =
Law of Conservation of Mass
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Chemical Equations
2 C(s) + O2(g) → 2 CO(g)
The “” symbol means “reaction proceeds in this direction” a “ “ symbol means the reaction is at equilibrium (s) = solid phase (l) = liquid phase (g) = gas phase (aq) = aqueous phase = heat
The 2’s are called “stoichiometric coefficients”
4 2
Stoichiometric Coefficients:
the ratios between reactants and/or products
1 mol O2
2 mol N2O5
2 mol N2O5
4 mol NO2
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Chapter Outline
7.1 Chemical reactions and the Conservation of Mass
7.2 Balancing Chemical Equations
7.3 Combustion Reactions
7.4 Stoichiometric Calculations and the Carbon Cyle
7.5 Percent Composition and Empirical Formulas
7.6 Empirical and Molecular Formulas Compared
7.7 Combustion Analysis
7.8 Limiting Reactants and Percent Yield
Guidelines for Balancing
Chemical Equations 1. Write an expression using correct chemical formulas for the
reactants and products, separated by an arrow (). Include
symbols indicating physical states.
2. For each element, add up the numbers of atoms on each
side. Check whether the expression is already balanced. If
so, you’re done!
3. Otherwise - if present - choose an element that appears in
only one reactant and product to balance first. Insert the
appropriate coefficient(s) to balance this element.
4. Choose the element that appears in the next fewest total
reactants and products and balance it. Repeat the process
for additional elements if necessary.
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1. Write an expression using correct chemical formulas for the
reactants and products, separated by an arrow (). Include
symbols indicating physical states.
2. For each element, add up the numbers of atoms on each
side. Check whether the expression is already balanced. If
so, you’re done!
Example 1: balance the reaction that occurs between nitrogen
dioxide and water to form nitric acid and nitrogen monoxide
NO2(g) + H2O(l) → HNO3(aq) + NO(aq)
N = 1 N = 2
O = 3 O = 4
H = 2 H = 1
3. Otherwise - if present - choose an element that appears in
only one reactant and product to balance first. Insert the
appropriate coefficient(s) to balance this element.
= H, so try a 2 in front of HNO3
NO2(g) + H2O(l) → HNO3(aq) + NO(aq)
NO2(g) + H2O(l) → 2 HNO3(aq) + NO(aq)
N = 1 N = 3
O = 3 O = 7
H = 2 H = 2
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4. Choose the element that appears in the next fewest total
reactants and products and balance it. Repeat the process
for additional elements if necessary.
= N, try a 3 in front of NO2
NO2(g) + H2O(l) → 2 HNO3(aq) + NO(aq)
N = 3 N = 3
O = 7 O = 7
H = 2 H = 2
3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(aq)
Done!
Chapter Outline
7.1 Chemical reactions and the Conservation of Mass
7.2 Balancing Chemical Equations
7.3 Combustion Reactions
7.4 Stoichiometric Calculations and the Carbon Cyle
7.5 Percent Composition and Empirical Formulas
7.6 Empirical and Molecular Formulas Compared
7.7 Combustion Analysis
7.8 Limiting Reactants and Percent Yield
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Combustion Reactions
The reaction of an organic compound with oxygen
to produce CO2 + H2O, for example, balance -
TIP FOR ALL COMBUSTION REACTIONS: Since oxygen
appears by itself, balance the other elements first, and then O2
CH4(g) + O2(g) → CO2(g) + H2O(g)
CH4(g) + O2(g) → CO2(g) + 2 H2O(g)
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
Done!
Example: C2H6 (ethane) is combusted in oxygen
Combustion Reactions
C2H6(g) + O2(g) → CO2(g) + H2O(g)
C2H6(g) + O2(g) → 2 CO2(g) + H2O(g)
C2H6(g) + O2(g) → 2 CO2(g) + 3 H2O(g)
C2H6(g) + 7/2 O2(g) → 2 CO2(g) + 3 H2O(g)
2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g)
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Chapter Outline
7.1 Chemical reactions and the Conservation of Mass
7.2 Balancing Chemical Equations
7.3 Combustion Reactions
7.4 Stoichiometric Calculations and the Carbon Cycle
7.5 Percent Composition and Empirical Formulas
7.6 Empirical and Molecular Formulas Compared
7.7 Combustion Analysis
7.8 Limiting Reactants and Percent Yield
Stoichiometric Calculations and the Carbon Cycle
Photosynthesis: 6 CO2(g) + 6 H2O(g) C6H12O6(aq) + 6 O2(g)
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Mauna Loa Observatory CO2 Concentrations since 1960 -
CO2 emissions over the last 800,000 years -
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1. Write the balanced chemical equation
2. Convert the mass of the reactant into moles
3. Use coefficients in the balanced equation to calculate the
number of moles of product (stoichiometric ratio)
4. Convert moles of products into grams (or other desired
quantities)
Amounts of Reactants & Products
= STOICHIOMETRY
1/MM ratio MM
Stoichiometry Example, p. 282
If the combustion of fossil fuels adds 8.2 x 1012 kilograms of
carbon to the atmosphere each year as CO2, what is the mass
of added carbon dioxide?
Step 1: Write the balanced chemical equation
Step 2: Convert quantities of known substances (C) into moles
C(s) + O2(g) CO2(g)
8.2 x 1012 kg C
1000 g C
kg x
1 mol C
12.0 g C x 6.83 x 1014 mol C =
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Stoichiometry Example, p. 282
Step 3: Use coefficients in balanced equation to calculate the
number of moles of CO2 (stoichiometric ratio)
Step 4: Convert moles of CO2 into grams
6.83 x 1014 mol C 1 mol CO2
1 mol C x = 6.83 x 1014 mol CO2
6.83 x 1014 mol CO2 44.0 g CO2
1 mol CO2
x = 3.00 x 1016 g CO2
Stoichiometry Example, p. 283
Suppose we wish to prepare 1.00 kg of acetylsalicylic acid. How
many grams of salicylic acid and how many grams of acetic
anhydride are needed?
C9H8O4
MW =
180.16
C4H6O3
MW =
102.09
C7H6O3
MW =
138.12
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Stoichiometry Example, p. 283
SA + AA ASA + Ac
1.00 kg ? g ? g
Step 1: Write the balanced chemical equation
Step 2: Convert quantities of known substances (ASA) into moles
1.00 kg ASA
1 mol ASA
180.16 g ASA x
1000 g ASA
kg ASA x = 5.551 mol ASA
Stoichiometry Example, p. 283
Step 3: Use coefficients in balanced equation to calculate the
number of moles of SA and AA (stoichiometric ratio)
mol SA = 5.551 mol ASA
mol AA = 5.551 mol ASA
1 mol SA
1 mol ASA x
1 mol AA
1 mol ASA x
= 5.551 mol SA
= 5.551 mol AA
SA + AA ASA + Ac
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Step 4: Convert moles of SA and AA into grams
Stoichiometry Example, p. 283
g SA = 5.551 mol SA
g AA = 5.551 mol AA
138.12 g SA
1 mol SA x
102.09 g AA
1 mol AA x
= 767 g SA
= 567 g AA
Sample Exercise 7.3 – Calculating the Mass of a
Product from the Mass of a Reactant.
Each year, power plants in the U.S. consume about 1.1 x 1011
kg of natural gas (CH4). How many kg of CO2 (MW = 44.01) are
released into atmosphere from these power plants. Given that
natural gas is mainly CH4 (MW = 16.04), base the calculation
on its combustion reaction –
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
1.1 x 1011 kg ? g
Step 1: Write the balanced chemical equation
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Step 2: Convert quantities of known substances (CH4) into moles
Sample Exercise 7.3 – Calculating the Mass of a
Product from the Mass of a Reactant.
1.1 x 1011 kg CH4 1 mol CH4
16.04 g CH4
x 1000 g CH4
kg CH4
x = 6.86 x 1012 mol
CH4
Step 3: Use coefficients in balanced equation to calculate the
number of moles of CO2 (stoichiometric ratio)
= 6.86 x 1012 mol CH4
1 mol CO2
1 mol CH4
x = 6.86 x 1012 mol CO2
Step 4: Convert moles of CO2 into grams
6.86 x 1012 mol CO2
44.01 g CO2
1 mol CO2
x = 3.0 x 1014 g CO2
Chapter Outline
7.1 Chemical reactions and the Conservation of Mass
7.2 Balancing Chemical Equations
7.3 Combustion Reactions
7.4 Stoichiometric Calculations and the Carbon Cycle
7.5 Percent Composition and Empirical Formulas
7.6 Empirical and Molecular Formulas Compared
7.7 Combustion Analysis
7.8 Limiting Reactants and Percent Yield
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Percent Composition: the composition of a compound in terms
of the percentage by mass of each element in the compound
n x molar mass of element molar mass of compound
x 100%
n is the number of moles of the element in 1 mole
of the compound
%H = 3 x (1.008 g)
97.99 g x 100% = 3.086%
%P = 1 x (30.97 g)
97.99 g x 100% = 31.60%
%O = 4 x (16.00 g)
97.99 g x 100% = 65.31%
3.086% + 31.60% + 65.31% = 99.96%
= 100%
H3PO4 MM = 97.99
g/mol
Sample Exercise 7.4 – Calculating Percent
Composition from a Chemical Formula
The mineral forsterite: Mg2SiO4,
MW = 140.71
%Mg = 2 x (24.31 g Mg)
140.71 g x 100% = 34.55%
%Si = 1 x (28.09 g Si)
140.71 g x 100% = 19.96%
%O = 4 x (16.00 g O)
140.71 g x 100% = 45.48%
34.55% + 19.96% + 45.48% = 99.99%
= 100%
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A formula showing the smallest whole number ratio of elements in a compound, e.g.
Benzene:
» Empirical = CH
» Molecular = C6H6
Glucose
» Empirical = CH2O
» Molecular = C6H12O6
Empirical Formula from % Composition
Empirical Formula from % Composition
1. Assume 100 g
2. Convert to moles
3. Divide by fewest number of moles
4. Convert the mole
ratio from step 3
into small whole
numbers if
necessary
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Sample Exercise 7.6
A sample of the carbonate mineral dolomite is 21.73% Ca,
13.18% Mg, 13.03% C, and the rest is oxygen. What is its
empirical formula?
Ca = 21.73 g
Mg = 13.18 g
C = 13.03 g
% O = 100 – 21.73 – 13.18 – 13.03 = 52.06 %
O = 52.06 g
1 mol Ca
40.078 g x = 0.54219 mol Ca
1 mol Mg
24.305 g x = 0.54228 mol Mg
1 mol C
12.011 g x = 1.0848 mol C
1 mol O
15.999 g x = 3.2540 mol O
1. Assume 100 g
2. Convert to moles
Sample Exercise 7.6
0.54219
0.54228
1.0848
3.2540
3. Divide by fewest number of moles
0.54219 Ca =
Mg =
C =
O =
0.54219
0.54219
0.54219
= 1.0
= 1.0
= 2.0
= 6.0
4. Convert the mole ratio from step 3 into small whole
numbers if necessary – NOT REQUIRED HERE
Empirical formula
= CaMgC2O6
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Example Illustrating Step 4
Vanillin is a common flavoring agent. It has a molar mass of 152 g/mol and is 63.15 %C and 5.30 %H; the rest is oxygen. What are the empirical and molecular formulas?
% O = 100 – 63.15 – 5.30 = 31.55 %
C = 63.15 g
H = 5.30 g
O = 31.55 g
1 mol C
12.011 g x = 5.258 mol C
1 mol H
1.008 g x = 5.258 mol H
1 mol O
15.999 g x = 1.972 mol O
1. Assume 100 g
2. Convert to moles
5.258
5.258
1.972
3. Divide by fewest number of moles
1.972 C =
H =
O =
1.972
1.972
= 2.67
= 2.67
= 1.0
4. Convert the mole ratio from step 3 into small whole numbers if necessary
Example Illustrating Step 4
Note that 2.67 = 2 2
3
So multiply by 3 which = 8
C =
H =
O =
2.67 x 3 = 8
1.0 x 3 = 3
2.67 x 3 = 8 Empirical formula =
C8H8O3
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Chapter Outline
7.1 Chemical reactions and the Conservation of Mass
7.2 Balancing Chemical Equations
7.3 Combustion Reactions
7.4 Stoichiometric Calculations and the Carbon Cycle
7.5 Percent Composition and Empirical Formulas
7.6 Empirical and Molecular Formulas Compared
7.7 Combustion Analysis
7.8 Limiting Reactants and Percent Yield
Empirical and Molecular
Formulas Compared
The molecular formula can be determined from the empirical
formula if the molecular weight of the compound is known.
glycolaldehyde
empirical = CH2O
formula weight (FW) = 30 g/mol
molecular = C2H4O2
Molecular weight (MW) = 60 g/mol
Note that molecular formula = empirical x 2
C2H4O2 = (CH2O) x 2 = C2H4O2
Or more generally –
molecular = (empirical) x R
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Empirical and Molecular
Formulas Compared
molecular = (empirical) x R it follows that -
MW = (FW) x R and therefore - R = MW
FW
e.g. glucose – assume we know the
MW = 180.0
Empirical = CH2O FW = 30.0
So molecular = (CH2O) x 6 = C6H12O6
R = 180.0
30.0 = 6
Molecular Mass and Mass
Spectrometry
Acetylene
C2H2
Benzene
C6H6
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Sample Exercise 7.7 – Using Percent Composition
and Molecular weight to Derive a Molecular Fomula
Pheromones are chemical substances secreted by members of a species
to stimulate a response in other individuals of the same species. The
percent composition of eicosene, a compound similar to the Japanese
beetle mating pheromone, is 85.63% C and 14.37% H. Its molecular mass,
as determined by mass spectrometry, is 280 amu. What is the molecular
formula of eicosene?
C = 85.63 g
H = 14.37 g
1 mol C
12.011 g x = 7.129 mol C
1 mol H
1.008 g x = 14.26 mol H
1. Assume 100 g
2. Convert to moles
7.129
14.26
3. Divide by fewest number of moles
7.129 C =
H = 7.129
= 1.00
= 2
4. Convert the mole ratio from step 3 into small whole numbers if necessary
– NOT NEEDED
Sample Exercise 7.7 – Using Percent Composition
and Molecular weight to Derive a Molecular Fomula
R = MW
FW
280
14 = 20 =
So molecular = (CH2) x 20 = C20H40
Empirical formula = CH2
FW = 14.0
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Chapter Outline
7.1 Chemical reactions and the Conservation of Mass
7.2 Balancing Chemical Equations
7.3 Combustion Reactions
7.4 Stoichiometric Calculations and the Carbon Cycle
7.5 Percent Composition and Empirical Formulas
7.6 Empirical and Molecular Formulas Compared
7.7 Combustion Analysis
7.8 Limiting Reactants and Percent Yield
Experimental Determination of Empirical
Formulas: Combustion Analysis
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Experimental Determination of Empirical
Formulas: Combustion Analysis
Calculation Outline:
CxHyOz + O2(g) x CO2(g) + y/2 H2O(g) excess mass
sample
Some of the oxygen comes
from the sample, some from
the excess O2
g CO2 mol CO2 mol C g C
g H2O mol H2O mol H g H
g of O = g of sample – g of C - g of H
Sample Exercise 7.8: Deriving an Empirical
Formula from Combustion Analysis Data
Combustion of 1.000 grams of an organic compound known to
contain only C, H and O produces 2.360 g CO2 and 0.640 g
H2O. What is the empirical formula of the compound?
mol C = 2.360 g CO2
12.011 g C
1 mol C x
1 mol CO2
44.0 g CO2 x
g C =
= 0.05364 mol C
0.05364 mol C
1 mol C
mol CO2 x
= 0.6442 g C
mol H = 0.640 g H2O
1.008 g H
1 mol H x
1 mol H2O
18.0 g H2O x
g H =
= 0.07111 mol H
= 0.07111 mol H
2 mol H
mol H2O x
= 0.07168 g H
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Sample Exercise 7.8: Deriving an Empirical
Formula from Combustion Analysis Data
g of O = g of sample – g of C - g of H
g of O = 1.000 g – 0.6442 g C - 0.07168 g H
g of O = 0.2841 g
and so the moles of O = 0.2841 g 1 mol O
15.999 g O x
= 0.01776 mol O
Sample Exercise 7.8: Deriving an Empirical
Formula from Combustion Analysis Data
0.05364 mol C
0.07111 mol H
0.01776 mol O
0.01776 C =
H =
O =
0.01776
0.01776
= 3
= 4
= 1
Empirical formula =
C3H4O
Now divide by the smallest number of moles -
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Chapter Outline
7.1 Chemical reactions and the Conservation of Mass
7.2 Balancing Chemical Equations
7.3 Combustion Reactions
7.4 Stoichiometric Calculations and the Carbon Cycle
7.5 Percent Composition and Empirical Formulas
7.6 Empirical and Molecular Formulas Compared
7.7 Combustion Analysis
7.8 Limiting Reactants and Percent Yield
Limiting Reagents
Limiting Reagents - a reactant that is consumed completely
in a chemical reaction before the other reactant(s) run out.
The amount of product formed depends on the amount of the
limiting reagent available.
Each sandwich consists of 2 slices bread, 1 slice of cheese, and 1 slice of salami
8 slices
bread
4 slices
cheese 3 slices
salami
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2 Br + 1 Ch + 1 Sal 1 sandwich
Limiting Reagents
8 4 3
Br in excess, so need
more salami! = LR
Stoichiometric ratio: 2 Br = 2.0
1 Sal
Given ratio: 8 Br = 2.7 tip: use a ratio > 1
3 Sal
Here’s the logic – compare the given number of moles to
the stoichiometric ratio
Given:
Strategy for determining which
reactant is the Limiting Reagent:
aA + bB C
then A is in excess and B is the limiting reagent
𝑚𝑜𝑙𝑒𝑠 𝐴
𝑚𝑜𝑙𝑒𝑠 𝐵𝑔𝑖𝑣𝑒𝑛
> 𝑎
𝑏 𝑠𝑡𝑜𝑖𝑐ℎ𝑖𝑜𝑚𝑒𝑡𝑟𝑖𝑐 if
1.Convert grams of each into moles
2.Calculate the stoichiometric ratio that’s the
largest, e.g. a/b or b/a
3.Calculate the given mole ratio in the same way
4.Compare to identify the LR
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Sample Exercise 7.9: Identifying the
Limiting Reagent in a Reaction Mixture
The flame in an acetylene torch reaches temperatures as high as 3500 oC as a result of the combustion of a mixture of acetylene (C2H2) and pure oxygen. If these two gases flow from high-pressure tanks at the rates of 52.0 g C2H2 and 188 g O2 per minute, which reactant is the limiting reagent, or is the mixture stoichiometric?
2 C2H2 + 5 O2 4 CO2 + 2 H2O
52.0 g 188 g
mol C2H2
Given ratio:
= 52.0 g C2H2
1 mol C2H2
26.0 g C2H2
x = 2.00 mol C2H2
mol O2 = 188 g O2
1 mol O2
32.0 g O2
x = 5.88 mol O2
2.00 mol C2H2
5.88 mol O2
= 2.94
tip: use a ratio > 1
Stoichiometric ratio:
Sample Exercise 7.9: Identifying the
Limiting Reagent in a Reaction Mixture
5 mol O2
2 mol C2H2 = 2.5
Given ratio:
= 2.94 2.00 mol C2H2
5.88 mol O2
therefore O2 is in excess and so C2H2 is the LR
𝑖𝑠 𝑚𝑜𝑙𝑒𝑠 𝑂2
𝑚𝑜𝑙𝑒𝑠 𝐶2𝐻2 𝑔𝑖𝑣𝑒𝑛
> 5 𝑚𝑜𝑙 𝑂2
2 𝑚𝑜𝑙 𝐶2𝐻2 𝑠𝑡𝑜𝑖𝑐ℎ𝑖𝑜𝑚𝑒𝑡𝑟𝑖𝑐
? ? ?
2 C2H2 + 5 O2 4 CO2 + 2 H2O
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Theoretical Yield is the maximum amount of product
formed from given quantities of reactants (check if a
limiting reagent is present).
Actual Yield is the amount of product actually obtained
from a reaction.
% Yield = Actual Yield
Theoretical Yield x 100
Percent Yield
Sample Exercise 7.10
The industrial process for making the ammonia used in fertilizer, explosives, and many other products is based on the reaction between nitrogen and hydrogen at high temperature and pressure.
If 18.2 kg of NH3 (MW = 17.03) is produced by a reaction mixture that initially contains 6.00 kg of H2 (MW = 2.016) and an excess of N2, what is the percent yield of the reaction?
N2(g) + 3 H2(g) → 2 NH3(g)
Actual yield =
18.2 kg
6.00 kg Excess
(no LR)
Calculate the theoretical yield of NH3 based on H2, and then calculate the %
yield -
% Yield = Actual Yield
Theoretical Yield x 100
g NH3 = 6.00 kg H2 kg
1000 g x
3 mol H2
2 mol NH3 x mol NH3
17.03 g NH3 x
= 3.379 x 104 g NH3
mol H2
2.016 g H2
x
= 33.79 kg NH3 = 33.8 kg NH3
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% Yield = Actual Yield
Theoretical Yield x 100
Sample Exercise 7.10
Actual Yield = 18.2 kg
Theoretical Yield = 33.79 kg
% Yield = 18.2 kg
33.79 kg x 100
= 53.8 %