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Starter S-125
1.2 moles NaC2H3O2 are used in a reaction. How many grams is that?
Stoichiometry
Chapter 12
12.1 The Arithmetic of Equations
Chapter 12
The basis for solving stoichiometry problems is a balanced chemical reaction
A balanced reaction is used to calculate
How much reactant is needed
How much product is produced
12.1 The Arithmetic of Equations
)(3)(2)(2 ggg NHHN )(3)(2)(2 23 ggg NHHN
Stoichiometry – the calculation of quantities
This reaction produces
Ammonia which is used
in fertilizers
Balanced reactions are usually
used to calculate grams
of product or reactant
12.1 The Arithmetic of Equations
)(3)(2)(2 23 ggg NHHN
Analysis of the reaction
Atoms – 2 atoms of nitrogen combine with 6 atoms of hydrogen – product is 2 nitrogen and 6 hydrogen
Molecules – 1 molecule of nitrogen gas combines with 3 molecules of hydrogen gas to produce 2 molecules of ammonia
12.1 The Arithmetic of Equations
)(3)(2)(2 23 ggg NHHN
Analysis of the reaction
Most useful
Moles – 1 mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia
12.1 The Arithmetic of Equations
)(3)(2)(2 23 ggg NHHN
Analysis of the reaction
Mass is always conserved in a chemical reaction
12.1 The Arithmetic of Equations
)(3)(2)(2 23 ggg NHHN
Analysis of the reaction
The key
moles can be converted to grams
grams can be converted to moles
Volume – remember that one mole of gas at STP is 22.4L
12.1 The Arithmetic of Equations
)(3)(2)(2 23 ggg NHHN
Analysis of the reaction
The key
moles can be converted to grams
grams can be converted to moles
Volume – remember that one mole of gas at STP is 22.4L
12.1 The Arithmetic of Equations
)(3)(2)(2 23 ggg NHHN
Analysis of the reaction
So if we start with 50g of N2, how many moles do we have
How many moles of NH3 would be produced?
For every 1 mole of N2, 2 moles of NH3
12.1 The Arithmetic of Equations
)(3)(2)(2 23 ggg NHHN
molg
molgN 8.1
28
150 2
3
2
32 6.31
28.1 molNH
molN
molNHmolN
Analysis of the reaction
How many grams of ammonia are produced
12.1 The Arithmetic of Equations
)(3)(2)(2 23 ggg NHHN
32
32 6.31
28.1 molNH
molN
molNHmolN
3
3
33 611
176.3 gNH
molNH
gNHmolNH
Starter S-126
A. Balance the following reactions
H2SO4 + KOH K2SO4 + H2O
B. What is the molar mass of the sulfuric acid.
C. If 20.0g of sulfuric acid is used, how many moles is that?
D. How many moles of water would be produced?
12.2 Chemical Calculations
Chapter 12
12.2 Chemical CalculationsMole Ratios
Come from balanced chemical reactions
Conversion factors derived from the coefficients in the balanced reactions
OHSOOSH 2222 OHSOOSH 2222 2232
2
2
2
3
molSO
molO2
2
2
2
molH O
molH S2
2
3
2
molO
molH S
12.2 Chemical Calculations
Mole-Mole Calculations
It is possible to convert from one quantity in a balanced reaction to another using mole ratios
For example – if 3.7 moles of sulfur dioxide is produced, how many moles of oxygen were used?
12.2 Chemical Calculations
2 2 2 22 3 2 2H S O SO H O
22 2
2
33.7 5.6
2
molOmolSO molO
molSO
Mass-Mass Calculations
Three steps
1. Convert given mass values into mole values.
2. Use a mole ratio to convert to the moles that the question requests
3. Convert this mole quantity to a mass value
12.2 Chemical Calculations
2 2 2 22 3 2 2H S O SO H O
Mass-Mass Calculations
How many grams of Oxygen are needed to produce 30.0g of Sulfur Dioxide?
12.2 Chemical Calculations
2 2 2 22 3 2 2H S O SO H O
22 2
2
130.0 0.468
64.06
molSOgSO molSO
gSO
2
2 22
30.468 0.702
2
molOmolSO molO
molSO
2
2 22
31.99880.702 22.5
1
gOmolO gO
molO
Volume-Mass Calculations
Same steps, but volume is converted to moles, or moles to volume
Example: If 4.0L of nitrogen monoxide reacts, how many grams of oxygen are used?
12.2 Chemical Calculations
( ) 2( ) 2( )g g gNO O NO ( ) 2( ) 2( )2 2g g gNO O NO
14.0 0.18
22.4
molNOLNO molNO
LNO
22
10.18 0.090
2
molOmolNO molO
molNO
22 2
2
31.99980.090 29
1
gOmolO gO
molO
Starter S-127
A. Balance the following reactions
Al(NO3)3 + Na2SO4 Al2(SO4)3+NaNO3
B. If 50.0g of Aluminum Nitrate reacts, how many grams of Aluminum Sulfate are produced?
12.3 Limiting Reagent and Percent Yield
Chapter 12
Limiting Reagent – the reactant that runs out first
Amounts of both reactants are given
Example: 80.0g Copper, 25.0g Sulfur
1. Calculate how many moles would each reactant produce
Copper
12.3 Limiting Reagent and Percent Yield
SCuSCu 2 SCuSCu 22
molCugCu
molCugCu 26.1
55.63
10.80
SmolCu
molCu
SmolCumolCu 2
2 630.02
126.1
Limiting Reagent – the reactant that runs out first
Amounts of both reactants are given
Example: 80.0g Copper, 25.0g Sulfur
1. Calculate how many moles would each reactant produce
Sulfur
12.3 Limiting Reagent and Percent Yield
SCuSCu 2 SCuSCu 22
molSgS
molSgS 780.0
06.32
10.25
SmolCu
molS
SmolCumolS 2
2 780.01
1780.0
Limiting Reagent – the reactant that runs out first
From the reactions
Copper would produce 0.630 mol Cu2S
Sulfur would produce 0.780 mol Cu2S
That means copper will run out first – it is the limiting reagent
Sulfur would be the excess reagent
12.3 Limiting Reagent and Percent Yield
SCuSCu 2 SCuSCu 22
One for you now
1. Balance the reaction
2. If 75.6g C2H4 reacts with 100.8g O2 – what is the limiting reagent?
Oxygen
3. How many moles of water are produced?
2.10 mol H2O
4. How many grams of water are produced?
12.3 Limiting Reagent and Percent Yield
OHCOOHC 22242 OHCOOHC 22242 223
4242
4242 70.2
05.28
16.75 HmolC
HgC
HmolCHgC
OmolH
HmolC
OmolHHmolC 2
42
242 40.51
270.2
2
2
22 15.3
00.32
18.100 molO
gO
molOgO
OmolHHC 242 40.5
OmolHmolO
OmolHmolO 2
2
22 10.23
215.3
OmolHO 22 10.2
OgHOmolH
OgHOmolH 2
2
22 8.37
1
02.1810.2
Starter S-132
A. Balance the following reactions
SiO2 + C SiC + CO
B. 35 grams of silicon dioxide reacts with 10.0g of Carbon, how much carbon monoxide is formed?
Reactions rarely produce as much product as is predicted-reactants can be
impure-reactions may not go
to completion-may compete with
smaller “side” reactions
In some reactions as little as 60% yield is considered a good result
12.3 Limiting Reagent and Percent Yield
Yield – how much product is produced
Theoretical Yield – the value calculated using stoichiometry
Actual Yield – the
amount of product
that actually forms
12.3 Limiting Reagent and Percent Yield
Percent Yield – a ratio of actual to theoretical yield
This number must be 100% or less
In lab the actual yield is the result you get
On a test, it will be a number that is given to you
The theoretical yield is calculated using limiting reactants
12.3 Limiting Reagent and Percent Yield
100%ActualYield
PercentYield xTheoreticalYield
Example: What is the theoretical yield of Calcium Oxide if 24.8g of Calcium Carbonate decomposes in the following reaction.
Balanced
12.3 Limiting Reagent and Percent Yield
3 2CaCO CaO CO
33 3
3
124.8 0.248
100.1
molCaCOgCaCO molCaCO
gCaCO
3
3
10.248 0.248
1
molCaOmolCaCO molCaO
molCaCO
56.10.248 13.9
1
gCaOmolCaO gCaO
molCaO
Example: What is the percent yield if actual yield is 9.6g?
12.3 Limiting Reagent and Percent Yield
56.10.248 13.9
1
gCaOmolCaO gCaO
molCaO
100%ActualYield
PercentYield xTheoreticalYield
9.6100%
13.9
gCaOPercentYield x
gCaO
69%PercentYield
Example: What is the theoretical yield if 15.0g of nitrogen reacts with 15.0g of hydrogen in the following reaction?
Balance
12.3 Limiting Reagent and Percent Yield
2 2 3N H NH
22 2
2
115.0 0.536
28.01
molNgN molN
gN
3
2 32
20.536 1.07
1
molNHmolN molNH
molN
2
2 22
115.0 7.44
2.016
molHgH moldH
gH
2 2 33 2N H NH 2 31.07N molNH
32 3
2
27.44 4.96
3
molNHmolH molNH
molH
2 34.96H molNH
33 3
3
17.0311.07 18.2
1
gNHmolNH gNH
molNH
Example: If the actual yield is 10.5g of NH3 what is the percent yield?
12.3 Limiting Reagent and Percent Yield
33 3
3
17.0311.07 18.2
1
gNHmolNH gNH
molNH
100%ActualYield
PercentYield xTheoreticalYield
3
3
10.5100% 57.7%
18.2
gNHx
gNH