Download Starter S-30

Starter S-125
1.2 moles NaC2H3O2 are used in a reaction.
How many grams is that?
Chapter 12
Stoichiometry
Chapter 12
12.1 The Arithmetic of Equations
12.1 The Arithmetic of Equations
The basis for solving stoichiometry problems
is a balanced chemical reaction
NN2 (2g()g 
3H 2 ( g )  2NH
NH3(3g()g )
) 
A balanced reaction is used to
calculate
How much reactant is
needed
How much product is
produced
12.1 The Arithmetic of Equations
Stoichiometry – the calculation of quantities
N 2 ( g )  3H 2 ( g )  2 NH 3( g )
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
Analysis of the reaction
N 2 ( g )  3H 2 ( g )  2 NH 3( g )
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
Analysis of the reaction
N 2 ( g )  3H 2 ( g )  2 NH 3( g )
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
Analysis of the reaction
N 2 ( g )  3H 2 ( g )  2 NH 3( g )
Mass is always conserved in a chemical
reaction
12.1 The Arithmetic of Equations
Analysis of the reaction
N 2 ( g )  3H 2 ( g )  2 NH 3( g )
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
Analysis of the reaction
N 2 ( g )  3H 2 ( g )  2 NH 3( g )
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
Analysis of the reaction
N 2 ( g )  3H 2 ( g )  2 NH 3( g )
So if we start with 50g of N2, how many
moles do we have
 2molNH
 1mol 3 
 2 
  1.38.mol
502gN
1.8molN
6molNH3
28 g2  
 1molN
How many moles of NH3 would be
produced?
For every 1 mole of N2, 2 moles of NH3
12.1 The Arithmetic of Equations
Analysis of the reaction
N 2 ( g )  3H 2 ( g )  2 NH 3( g )
How many grams of ammonia are produced
 2molNH
 
17 gNH
3 3
 
 3.661
13.8.6molN
molNH
molNH
gNH 33
2 3
1molNH
2 3 
 1molN
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?
Chapter 12
12.2 Chemical Calculations
12.2 Chemical
Chemical Calculations
Calculations
12.2
Mole Ratios
Come from balanced chemical reactions
2HH
 3O  2SO  2H O
2S
2 S  O22  SO2 2 H 2O2
Conversion factors derived from the coefficients in
the balanced reactions
2molSO2 2molH 2O 3molO2
3molO2 2molH 2 S 2molH 2 S
12.2 Chemical Calculations
Mole-Mole Calculations
2H 2 S  3O2  2SO2  2H 2O
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?
 3molO2 
3.7molSO2 
  5.6molO2
 2molSO2 
12.2 Chemical Calculations
Mass-Mass Calculations
2H 2 S  3O2  2SO2  2H 2O
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
Mass-Mass Calculations
2H 2 S  3O2  2SO2  2H 2O
How many grams of Oxygen are needed to
produce 30.0g of Sulfur Dioxide?
 1molSO
 
31.9988
3molO
2 2gO2
0.468
30.0
0.702
gSO
molSO
molO
 22.5
molSO
molO
gO222
  0.468
0.702
2  22 
1gSO
molO
 64.06
 2molSO
2 
22  
12.2 Chemical Calculations
Volume-Mass Calculations
2 NO( g )  O2(2(gg))  2NO
NO2(2(g )g )
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?
131.9998
molO2  gO2 
 1molNO
0.090
molO
29
gO2 2
4.0molNO
LNO
0.18
0.090
molO
 molNO
 2 1molO
 0.18
LNO  2 
 22.4
2molNO
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?
Chapter 12
12.3 Limiting Reagent and Percent Yield
12.3 Limiting Reagent and Percent Yield
Limiting Reagent – the reactant that runs out
first
2Cu
CuSS
Cu
Cu22SS
Amounts of both reactants are given
Example: 80.0g Copper, 25.0g Sulfur
1. Calculate how many moles would each
reactant produce
Copper
1molCu
1molCu
S 

2
.0 gCu
.26molCu 2 S
1.2680
molCu
 
 01.630
.55 gCu 
molCu
  263
12.3 Limiting Reagent and Percent Yield
Limiting Reagent – the reactant that runs out
first
2Cu
CuSS
Cu
Cu22SS
Amounts of both reactants are given
Example: 80.0g Copper, 25.0g Sulfur
1. Calculate how many moles would each
reactant produce
Sulfur
1molS2 S 
1molCu

  0.780molS
.0 gS 
0.78025
molS
molCu2 S
.06 gS 
1molS
  32
12.3 Limiting Reagent and Percent Yield
Limiting Reagent – the reactant that runs out
first
2Cu
CuSS
Cu
Cu22SS
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
One for you now
CC
3O22  2CO
CO
22H
O2O
2H
2H
4 4
2 2H
1. Balance the reaction
2. If 75.6g C2H4 reacts with 100.8g O2 – what
is the limiting reagent?
C2 H 4  5.40molH 2O
O2  2.10molH 2O
Oxygen  12
 
molC
12
molO
molH
18
molH
.02
O
2H
22gH
24O2O
2 2moles
O  of water

2How
.70
.32100
6.15
gC
10
molC
.molO
molH
82 H
gO
32are
.15
2.5
10
..70
40
molO
37
molH
molC
molH
.8 gH
O4
24 H
24
2 22O
2H
3. 75
many
produced?
.molC
00
molO
1gC
molH
gO22H
2H424O
 28
32
 .1
305
 
2.10 mol H2O
4. How many grams of water are produced?
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?
12.3 Limiting Reagent and Percent Yield
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
 ActualYield
PercentYield  
 TheoreticalYield

 x100%

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
Example: What is the theoretical yield of
Calcium Oxide if 24.8g of Calcium
Carbonate decomposes in the following
reaction.
CaCO3  CaO  CO2
Balanced
56.1
gCaO
 1molCaCO
 1
 
molCaO
3
0.248
molCaO
gCaO
24.8
0.248
gCaCO
molCaCO

13.9
0.248
molCaCO
molCaO3
 0.248

3  3 
1
molCaO
gCaCO
3 3 
 100.1
 1molCaCO
12.3 Limiting Reagent and Percent Yield
Example: What is the percent yield if actual
yield is 9.6g?
9.6 gCaO  
 ActualYield
PercentYield
  
PercentYield
PercentYield
 69% x100%
 x100%
 TheoreticalYield
 13.9 gCaO  
 56.1gCaO 
0.248molCaO 
  13.9 gCaO
 1molCaO 
12.3 Limiting Reagent and Percent Yield
Example: What is the theoretical yield if
15.0g of nitrogen reacts with 15.0g of
hydrogen in the following reaction?
NN
3H 2  2NH
NH
2 2
33
Balance
N 2  1.07molNH 3
H 2  4.96molNH 3
 11217.031


2
molNH
molH
molN


molNH
3
22gNH
3
3
0.536
15.0
15.0
gH
molN
gN22232

0.536
1.07
moldH
molN
molNH
1.07
7.44
molNH
molH
4.96
 18.2
molNH
gNH
 7.44
2 333

molN
gH
gN2222 3 
molH
molNH
2.016
 311
28.01
12.3 Limiting Reagent and Percent Yield
Example: If the actual yield is 10.5g of NH3
what is the percent yield?
 10.5 gNH
 3 ActualYield 
PercentYield
   x100%  57.7%

 x100%
 TheoreticalYield

3 
 18.2 gNH
 17.031gNH 3 
1.07molNH 3 
  18.2 gNH 3
 1molNH 3 