1. Gas Stoichiometry Review

2. Gas Stoichiometry
Many chemical reactions involve gases as a reactant or a product
Gas Stoichiometry – the procedure for calculating the volume of gases as products or reactants
Gases have a molar volume (L/mole) rather than concentration.
This is the conversion factor used to convert (litres of gas) to (moles of gas)
The Ideal Gas Law (PV = nRT) may also be required to:
A) find the number of moles of reactant
B) Find the V, P, or T of the product

3. The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).
Experiments show that at STP, 1 mole of an ideal gas occupies L.
PV = nRT
R = PV nT =
(1 atm)(22.414L)
(1 mol)( K)
R = L • atm / (mol • K)

4. PV = nRT nRT V = P 1.37 mole x 0.0821 x 273.15 K V = 1 atmo V = 30.6 L
What is the volume (in liters) occupied by 49.8 g of HCl at STP?
T = 0 0C = K
P = atmo
PV = nRT
n = 49.8 g x
1 mole HCl
36.45 g HCl
= 1.37 mole
V =
nRT P
V =
1 atmo
1.37 mole x x K
L•atmo
mol•K
V = 30.6 L

5. PV = nRT n, V and R are constant P1 T1 P2 T2 = P2 = P1 x T2 T1
Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atmo and 18 oC is heated to 85 oC at constant volume. What is the final pressure of argon in the lightbulb (in atmo)?
PV = nRT
n, V and R are constant
P1 = 1.20 atm
T1 = 291 K
P2 = ?
T2 = 358 K P1 T1 P2 T2 =
P2 = P1 x T2 T1
= 1.20 atmo x
358 K
291 K
= 1.48 atmo

6. Example
Hydrogen gas is produced when sodium metal is added to water. What mass of sodium is necessary to produce 20.0L of hydrogen at STP?
m? L
2Na(s) H2O (l)  2NaOH(aq) + H2(g)
20.0L x 1 mole = mole H2 L
mole H2 x 2 mole Na = mole Na
1 mole H2
THEN mole Na x g Na = g Na
1 mole
= 41.0 g Na

7. Example
If the conditions are not STP, the molar volume cannot be used! You must use the ideal gas law to find the gas values using moles determined from stoichiometry
What volume of ammonia at 450kPa and 80oC can be obtained from the complete reaction of 7.5kg of hydrogen with nitrogen?
7500 g V ?
2N2(g) H2(g)  NH3(g)
7500 g x 1 mole H2 = mole H2 g
mole H2 x 2 mole NH3 = mole NH3
3 mole H2
Also: 450 kPa x 1 atmo = atmo and 80oC = K
kPa
PV = nRT  V = nRT P
V = ( mole)( Latmo/moleK)(353.15K)
( atmo)
= L  x 104 L of NH3(g)

8. Gas Stoichiometry Summary
Write a balanced chemical equation and list the measurements, unknown quantity symbol, and conversion factors for the measured and required substances.
Convert the measured quantity to moles using the appropriate conversion factor
Calculate the amount of moles of the required substance using the mole ratio from the balanced chemical equation.
Convert the calculated moles to the final quantity requested using the appropriate conversion factor.

9. Gas Stoichiometry
Be careful! You CANNOT always use mole/ L

10. C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)
Gas Stoichiometry
What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l)
g C6H12O mole C6H12O mole CO V CO2
1 mole C6H12O6
g C6H12O6 x
5.60 g C6H12O6
= mole C6H12O6
6 mole CO2
1 mole C6H12O6
mole x
= mole CO2
0.187 mole x x K
L•atm
mol•K
1.00 atm =
nRT P
V =
= 4.76 L