ch 4: using balanced chemical equations mole/mole...
TRANSCRIPT
10/7/2009 1
Chem 105Wednesday, 7 Oct 2009
Ch 4: Using balanced chemical equations
Mole/moleMole/massStoichiometry and the 1947 Texas City Disaster
Start with clicker question on balancing chem rxn
10/7/2009 2
Write out the correct balanced chemical equation for the
combustion of C5H12.
Now, INCLUDING “1” if necessary, what is the SUM of all
stoichiometric coefficients?...
17 20 21 22
11
67
50
1. 17
2. 20
3. 21
4. 2232
42 (57%)
10/7/2009 3
What is the SUM of the coefficients in the balanced
chemical equation for the combustion of C5H12 to give
water and carbon dioxide?
1. seventeen
2. twenty
3. twenty-one
4. twenty-two
Start: C5H12 + O2 ---> CO2 + H2O
Bal C: C5H12 + O2 ---> 5CO2 + H2O
Bal H: C5H12 + O2 ---> 5CO2 + 6H2O
Bal O: C5H12 + 8O2 ---> 5CO2 + 6H2O
Not balanced
Not balanced
Not balanced
Balanced!
10/7/2009 4
Using balanced chemical equations in stoichiometric
calculations
Answers questions like “How much product is formed in a
given chemical reaction?”
Or “How much reactant B does it require for reactant A to
react completely?”
Key to success is to always use the coefficients of the
balanced chemical equation to relate amounts of reactants
and products.
10/7/2009 5
C5H12 (l) + 8 O2 (g) --> 5 CO2 (g) + 6 H2O (g)
This equation contains several useful ratios:
“8 moles of O2 are required for complete combustion
of 1 mole of C5H12”
“Complete combustion of 1 mole of C5H12 gives 6
moles of water vapor.”
“Complete combustion of 10 moles of C5H12 gives 60 moles of water vapor.”
10/7/2009 6
How many moles of C5H12 are required to produce 1 mole of carbon
dioxide per the combustion reaction shown?...
C5H12 (l) + 8 O2 (g) --> 5 CO2 (g) + 6 H2O (g)
Because this is a “chemical” equation, the items on the left EQUAL the items on the right. So we can use conversion factors to convert between these different amounts.
answerfinalwstartyouwhat ?=
×
852 ?1 HCmolCOmol =
×
8585
2 ?1 HCmolHCmol
COmol =
×
85
2
852 ?1 HCmol
COmol
HCmolCOmol =
×
85
2
852 ?
5
11 HCmol
COmol
HCmolCOmol =
×
85
2
852 2.0
5
11 HCmol
COmol
HCmolCOmol =
×
10/7/2009 7
OWL problems
Mole – mole ����
10/7/2009 8
We also use the balanced chemical equation for solving mole/mass, mass/mass problems.
We use the “Mole City Analogy” to guide our thinking.
10/7/2009 9
Mole St.
“Mole City”Analogy, Prof. Addison Ault, Cornell College
Gram St.
Mole St.
Gram St.
Mo
lar
Ma
ss A
ve
W
Mo
lar
Ma
ss A
ve
E
To go from the Product side of Mole City from the Reactant side,
you must cross Mole St. Bridge.
Mo
lari
tyA
ve
W
Volume of Solution St.
Mo
lari
tyA
ve
E
Volume Pure Liq St. Volume Pure Liq St.
Mole St. Bridge
Vol Solution St.
De
nsity A
ve
. W
De
nsity A
ve
. E
Mole
Ri v
er
Reactant side Product side
10/7/2009 10
A given amount of reactant �
gives what is mass of product?
This requires the use of the balanced chemical
equation and its mole coefficients.
10/7/2009 11
Mole St.
Gram St.
Mole St.
Gram St.
Mo
lar
Ma
ss A
ve
W
Mo
lar
Ma
ss A
ve
E
“Reactant” side of city “Product” side of city
Mo
lari
tyA
ve
W
Volume of Solution St.
Mo
lari
tyA
ve
E
Volume Pure Liq St. Volume Pure Liq St.
Mole St.Bridge
Vol Solution St.
De
nsity A
ve
. W
De
nsity A
ve
. E
No
10/7/2009 12
What mass (g) of water is produced by combustion of 2.5 moles of C5H12?
C5H12 (l) + 8 O2 (g) --> 5 CO2 (g) + 6 H2O (g)
10/7/2009 13
Mole St.
Gram St.
Mole St.
Gram St.
Mo
lar
Ma
ss A
ve
W
Mo
lar
Ma
ss A
ve
E
“Reactant” side of city “Product” side of city
Mo
lari
tyA
ve
W
Volume of Solution St.
Mo
lari
tyA
ve
E
Volume Pure Liq St. Volume Pure Liq St.
Mole St.Bridge
Vol Solution St.
De
nsity A
ve
. W
De
nsity A
ve
. E
10/7/2009 14
What mass (g) of water is produced by combustion of 2.5 moles of C5H12?
C5H12 (l) + 8 O2 (g) --> 5 CO2 (g) + 6 H2O (g)
OHg2.7x10OHg0.227OHmol
OHg18.016
HCmol1
OHmol6HCmol2.5
OHg?OHmol
OHg18.016
HCmol1
OHmol6HCmol2.5
OHg?OHmol
OHg
HCmol
OHmolHCmol2.5
OHg?OHg
HCmolHCmol2.5
OHg?OHg
HCmol2.5
OHg?HCmol2.5
answer?datastarting
2
2
2
2
2
125
2125
2
2
2
125
2125
2
2
2
125
2125
22
125
125
22
125
2125
==
×
=
×
=
×
=
×
=
×
=
×
=
×
Coefficients of
balanced chem equnMolar mass of water
10/7/2009 15
Mole St.
Gram St.
Mole St.
Gram St.
Mo
lar
Ma
ss A
ve
W
Mo
lar
Ma
ss A
ve
E
“Reactant” side of city “Product” side of city
Mo
lari
tyA
ve
W
Volume of Solution St.
Mo
lari
tyA
ve
E
Volume Pure Liq St. Volume Pure Liq St.
Mole St.Bridge
Vol Solution St.
De
nsity A
ve
. W
De
nsity A
ve
. E
10/7/2009 16
OWL problems
Limiting reagent When 2 or more chemicals react to form products, the limiting
reagent (or limiting reactant) = the one that, if converted
completely into products, give the smaller amount of product.
Or another way to look at it is, if all the
reactants present react, then all the limiting
reactant is used up. The excess reactant
is the one for which some remains un-reacted.
These terms are defined for particular
situations, where certain amounts of
chemicals are mixed together in aqueous solution, or as gases, as shown here.
10/7/2009 17
How many moles of water are produced by
decomposition of 6.35 x 105 kg of ammonium
nitrate by the following reaction?
NH4NO3 (s) --> 2H2O (g) + N2O (g)
Texas City Disaster 1947
10/7/2009 18
Texas City Disaster of 1947:
Decomposition of ammonium nitrate
Two shiploads of ammonium nitrate (used as agricultural fertilizer) exploded,
killing 541 people. The ships contained 1.40 x 106 lbs (6.35 x 105 kg) of
ammonium nitrate.
One of the reactions that occurred was NH4NO3 (s)--> N2O (g) + 2H2O (g)
10/7/2009 19
10/7/2009 20
http://www.local1259iaff.org/disaster.html
It is estimated that more than 100 km3 of N2O gas and water vapor were
produced by the explosion. The shockwave was felt in Austin 200 miles away.
10/7/2009 21
How many moles of water vapor can be produce by reaction of
6.35 x 105 kg of ammonium nitrate?
NH4NO3 (s)--> N2O (g) + 2H2O (g)
10/7/2009 22
Mole St.
Gram St.
Mole St.
Gram St.M
ola
r M
ass A
ve
W
Mo
lar
Ma
ss A
ve
E
“Reactant” side of city “Product” side of city
Mo
lari
tyA
ve
W
Volume of Solution St.
Volume Pure Liq St. Volume Pure Liq St.
Mole St.Bridge
Vol Solution St.
De
nsity A
ve
. W
De
nsity A
ve
. E
Kilogram St.M
etr
ic C
on
ve
rsio
n A
ve
.
10/7/2009 23
OHmolANmol1
OHmol2
AN80.04g
ANmol1
ANkg1
ANg10ANkg10x6.35
OHmolANmol
OHmol
ANg
ANmol
ANkg
ANgANkg10x6.35
22
35
225
=
×
=
×
These factors (1)
convert kg AN to g AN,
then (2) convert g AN
to moles AN using the molar mass of AN, and
finally (3) convert
moles AN to moles of
water using the
coefficients of balanced chemical
equation.
Build in UNITS first. Start with a notion about what units survive in the answer, and what units in the starting
data must be canceled.
In this case, “mol H2O” must survive, and “kg AN” must be canceled.
OHmol...ANkg10x6.35
answer...datastarting
2
5=
×
=
×
OHmolANmol
OHmol
ANkg
ANgANkg10x6.35
OHmolOHmol
ANkgANkg10x6.35
225
225
=
×
=
×
OHmol10x1.59
OHmol10x671.58
2
7
2
7
= NH4NO3 (s)--> 2H2O (g) + N2O (g)
10/7/2009 24
These calculations may be done stepwise by
calculating the intermediate quantities.
However:
DO NOT ROUND OFF INTERMEDIATE
QUANTITIES.
Round-off errors lose points.
10/7/2009 25
Hint: To avoid round-off error, also use exact atomic and molecular weights, with at least 2 more significant figures than your data.
10/7/2009 26
The above reaction can be used (cautiously!) to make dinitrogen monoxide
in the lab, too. How many grams of ammonium nitrate would be needed to
make 10. g of N2O?
1. A
2. B
3. C
4. D
80.4
1
1
2
1
44.010:B
1
80.4
1
2
1
44.010:C
1
80.4
1
2
44.0
110:D
1
80.4
1
1
44.0
110:A
NH4NO3 (s) --> N2O (g) + 2H2O (g)
10/7/2009 27
1. A
2. B
3. C
4. D
80.4
1
1
2
1
44.010:B
1
80.4
1
2
1
44.010:C
1
80.4
1
2
44.0
110:D
1
80.4
1
1
44.0
110:A
The above reaction can be used (cautiously!) to make dinitrogen monoxide. How many grams of ammonium nitrate are needed to make 10. g of N2O?
NH4NO3 (s) --> 2H2O (g) + N2O (g)
ANg18ANg.2718ANmol1
ANg80.4
ONmol1
ANmol1
ON44.0
ONmol1ONg10.
22
22 ==
×
g