1. Bellwork:
What is the volume, in liters, of mol of oxygen gas at 20.0ºC and atm pressure?
V = ?
n = mol
T = 20ºC = 293 K
P = atm
PV = nRT
6.17 L O2
nRT P
0.250 mol (.0821 L·atm/mol·K) (293 K)
0.947 atm = V = =

2. Gas Stoichiometry at Non-STP Conditions

3. OBJECTIVES
Use a chemical equation to specify volume ratios for gaseous reactants or products, or both.
Use volume ratios and the gas laws to calculate volumes, masses, or molar amounts of gaseous reactants or products.

4. Gas Stoichiometry Moles  Liters of a Gas: Non-STP
STP - use 22.4 L/mol
Non-STP - use ideal gas law
Non-STP
Given liters of gas?
start with ideal gas law
Looking for liters of gas?
start with stoichiometry conv.
If given L of gas and are looking for L of a different gas, just use the mole ratio!

5. Gas Stoichiometry Problem
What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC?
CaCO3  CaO CO2
5.25 g
? L non-STP
Looking for liters: Start with stoich and calculate moles of CO2.
5.25 g
CaCO3
1 mol
CaCO3
100.09g
1 mol
CO2
CaCO3
= mol CO2
Plug this into the Ideal Gas Law to find liters.

6. Gas Stoichiometry Problem
What volume of CO2 forms from g of CaCO3 at 103 kPa & 25ºC?
GIVEN:
P = 103 kPa
V = ?
n = mol
T = 25°C = 298 K
R = LkPa/molK
WORK:
PV = nRT V = nRT/P
V=(0.0525mol)(8.315LkPa/molK)(298K)
(103 kPa)
V = 1.26 L CO2

7. Gas Stoichiometry Problem
How many grams of Al2O3 are formed from L of O2 at 97.3 kPa & 21°C?
4 Al O2  2 Al2O3
15.0 L
non-STP
? g
GIVEN:
P = 97.3 kPa
V = 15.0 L
n = ?
T = 21°C = 294 K
R = LkPa/molK
WORK:
PV = nRT
n = PV/RT
n= (97.3 kPa) (15.0 L) (8.315LkPa/molK) (294K)
n = mol O2
Given liters: Start with Ideal Gas Law and calculate moles of O2.
NEXT 

8. Gas Stoichiometry Problem
How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?
4 Al O2  2 Al2O3
15.0L
non-STP
? g
Use stoich to convert moles of O2 to grams Al2O3.
0.597
mol O2
2 mol Al2O3
3 mol O2
g Al2O3
1 mol
Al2O3
= 40.6 g Al2O3

9. CaCO3(s) CaO(s) + CO2(g)
? g
5.00 L
How many grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide gas at STP?
V = 5.00 L
T = 0ºC = 273 K
P = 1 atm
n = ?
At STP
1 mole = 22.4 L
5.00 L n
1 mol
22.4 L
.223 mol CO2 = x =
1 mol CaCO3
1 mol CO2
g CaCO3
1 mol CaCO3
22.3 g CaCO3
.223 mol CO3 = x x

10. WO3(s) + 3H2(g) W(s) + 3H2O(g)
? L
How many liters of hydrogen gas at 35ºC and atm are needed to react completely with 875 g of tungsten oxide?
1 mol WO3
g WO3
3 mol H2
1 mol WO3
11.3 mol H2 =
875 g WO3 x x
n = mol
P = atm
T = 35ºC = 308 K
V = ?
PV = nRT
nRT P
11.3 mol (.0821 L·atm/mol·K) (308 K)
(0.980 atm) V =
292 L H2 = =