Transcript Gas Stoichiometry - Derry Area School District

Gas Stoichiometry
Relationships
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2 H2 + O2  2 H2O
2 molecules + 1 molecule
2 moles
+ 1 mole
4 atoms
+ 2 atoms
4 grams
+ 32 grams
2[6.02 x 1023] + 1[6.02 x 1023]
particles
particles
yields
yields
yields
yields
yields
2 molecules
2 moles
6 atoms
36 grams
2[6.02 x 1023]
particles
Recalling Avagadro’s Principle
 equal volumes of gases [at the same temp. and
press.] contain equal numbers of molecules,
 Gaseous reactions occuring at the same
temperature and pressure have a volumetric
relationship
2 H2 + O2  2 H2O
 2 volumes H2+ 1 volume O2  2 volumes H2O
vapor
Joseph Gay-Lussac
 First chemist to apply this idea to chemical
reactions involving gases.
 Law of Combining Gas Volumes
 [At the same temp. and press.] the volumes
of reacting gases and their gaseous
product[s] are expressed in small whole
number ratios.
 coefficients
Using Gay-Lussac’s idea, Avagadro
was able to identify the 7 diatomic
elements.
 At S.T.P.
 contains 1 mole
22.4 L
 6.02 x 1023 chemical units
of any gas
 O mass should equal 16 g
 O mass actually equals 32 g
 Oxygen must be diatomic (O2)
•Since molar ratios are the same as
volume ratios - the liter volumes can
be used in the same manner as the
molar ratios in calculations!
Have ratio will be the same for
Want
mole ratios and volume ratios (when
all of the substances are gases).
Volume - Volume Calculations
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FACT: Air is 20.9 % oxygen
Based on the equation
2C8H18[g] + 25O2[g]  16CO2[g] +18H2O[g]
A. How many liters of AIR must enter the
carburetor to complete the combustion of 30.0 L of
octane, C8H18?
 B. How many liters of CO2 are formed?
 Assume all measurements are at the same temp.
and press.
 2C8H18[g] + 25O2[g]  16CO2[g] +18H2O[g]
 30 L
x
 30 L C8H18 | 25 L O2 = 375 L O2
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| 2 L C8H18
 Since air is 20.9 % [.209] oxygen
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375 L O2 = [.209]x
1794 L air = x
 2C8H18[g] + 25O2[g]  16CO2[g] +18H2O[g]
 30.0 L
x
 30 L C8H18 | 16 L CO2 = 240 L CO2
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| 2 L C8H18
Mass - Volume Calculations
 How many grams of CaCO3 must
decompose to produce 5.0 L of CO2 at
S.T.P.?
 CaCO3[s]  CaO[s] + CO2[g]
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x
5L
 *CaCO3 is chalk, a solid.
 Gay-Lussac’s Law applies only for chemical
reactions in which all components are
gaseous.!!!!!!!!!!!!!!!
What do we know?:
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* 1 mole of CO2 weighs 44 g
* 1 mole of CaCO3 weighs 100 g
* 1 mole of CO2 occupies 22.4 L at
S.T.P.
Assuming S.T.P.
 CaCO3[s]  CO2[g] +CaO[s]
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x
5L
 5 L CO2 | 1 mol CaCO3 | 100 g CaCO3 | 1 mol CO2 = 22.3 g Ca CO3
| 1 mol CO2
| 1 mol CaCO3| 22.4 L
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 STOICHIOMETRY
MOLAR VOLUME
STEP
Try this one!
 What volume of O2, collected by water
displacement, at 20oC and 750 Torr, can be
obtained by the decomposition of 55.0 g of
KClO3?
2KClO3  2 KCl + 3O2
55 g
x
Strategy
2KClO3  2 KCl + 3O2
55 g
x
 First solve as if at S.T.P.
55g KClO3 | 1 mol KClO3 | 3 mol O2
| 22.4 L O2 = 15.1 L @ S.T.P.
| 122.5 g KClO3 | 2 mol KClO3 | 1 mol O2
Strategy
 Now change to the conditions in the
question.
 750 Torr - 17.5 Torr =
Vapor pressure
of water at 20oC
732.5 Torr
Pressure of the dry gas
Strategy
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Using the Combined Gas Law to find the new volume.
V1 = 15.1 L
V2 = x
T1 = 273 K
T2 = 293 K
P1 = 760 Torr
P2 = 732.5 Torr
V1P1 = V2P2
T1
T2
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[15.1 l][760 Torr] = x[732.5 Torr]
273 K
293 K
x = 16.9 L
With Limiting Reactants
 What volume of CO2 is produced at S.T.P. from
4.14 g Ca3[PO4]2 and 1.20 g SiO2?
 2Ca3[PO4]2[cr]
1.34 x 10-2 mol
+ 6SiO2[cr]
10C[amor] + 6CaSiO3[cr] + 10CO2[g] + P4[cr]
2.0 x 10-2 mol
 Determine the L.R.
– x = 1.5 L, using Ca3[PO4]2
– x = .76 L, using SiO2 , therefore SiO2 is the L.R.
x
And the answer is…
 2.0 x 10-2 mol SiO2 | 10 mol CO2 | 22.4 L CO2 = .747 L CO2
| 6 mol SiO2 | 1 mol CO2