Transcript IV. Gas Stoichiometry (non-STP)

Ch. 10 & 11 - Gases
IV. Gas Stoichiometry at
Non-STP Conditions
(p. 347-350)
C. Johannesson
A. Gas Stoichiometry
 Liters of a Gas:
• STP - use 22.4 L/mol
• Non-STP - use ideal gas law
 Moles
 Non-STP
• Given liters of gas?
 start with ideal gas law
• Looking for liters of gas?
 start with stoichiometry conv.
C. Johannesson
B. Gas Stoichiometry Problem
 What
volume of CO2 forms from 5.25 g
of CaCO3 at 103 kPa & 25ºC?
CaCO3
5.25 g

CaO
+
Looking for liters: Start with stoich
and calculate moles of CO2.
5.25 g 1 mol
CaCO3 CaCO3
1 mol
CO2
100.09g 1 mol
CaCO3 CaCO3
C. Johannesson
CO2
?L
non-STP
= 1.26 mol CO2
Plug this into the Ideal
Gas Law to find liters.
B. Gas Stoichiometry Problem
 What
volume of CO2 forms from
5.25 g of CaCO3 at 103 kPa & 25ºC?
GIVEN:
WORK:
P = 103 kPa
V=?
n = 1.26 mol
T = 25°C = 298 K
R = 8.315 dm3kPa/molK
PV = nRT
(103 kPa)V
=(1mol)(8.315dm3kPa/molK)(298K)
V = 1.26 dm3 CO2
C. Johannesson
B. Gas Stoichiometry Problem
 How
many grams of Al2O3 are formed from
15.0 L of O2 at 97.3 kPa & 21°C?
4 Al
+
3 O2

15.0 L
non-STP
2 Al2O3
?g
GIVEN:
WORK:
P = 97.3 kPa
V = 15.0 L
n=?
T = 21°C = 294 K
R = 8.315 dm3kPa/molK
PV = nRT
(97.3 kPa) (15.0 L)
= n (8.315dm3kPa/molK) (294K)
Given liters: Start with
Ideal Gas Law and
calculate moles of O2.
NEXT 
n = 0.597 mol O2
C. Johannesson
B. Gas Stoichiometry Problem
 How
many grams of Al2O3 are formed
from 15.0 L of O2 at 97.3 kPa & 21°C?
3 O2 
15.0L
Use stoich to convert moles
of O to grams Al O .
non-STP
0.597 2 mol 101.96 g
mol O2 Al2O3
Al2O3
4 Al
2
2
+
2 Al2O3
?g
3
3 mol O2
1 mol
Al2O3
C. Johannesson
= 40.6 g Al2O3