1. Gas Stoichiometry

2. 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 conversion
Courtesy Christy Johannesson

3. 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
= 1.26 mol CO2
Plug this into the Ideal Gas Law to find liters.
Courtesy Christy Johannesson

4. Gas Stoichiometry Problem
What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC?
GIVEN:
P = 103 kPa
V = ?
n = 1.26 mol
T = 25°C = 298 K
R = dm3kPa/molK
WORK:
PV = nRT
(103 kPa)V =(1mol)(8.315dm3kPa/molK)(298K)
V = 1.26 dm3 CO2
Courtesy Christy Johannesson

5. 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
15.0 L
non-STP
? g
GIVEN:
P = 97.3 kPa
V = 15.0 L
n = ?
T = 21°C = 294 K
R = dm3kPa/molK
WORK:
PV = nRT
(97.3 kPa) (15.0 L) = n (8.315dm3kPa/molK) (294K)
n = mol O2
Given liters: Start with Ideal Gas Law and calculate moles of O2.
NEXT 
Courtesy Christy Johannesson

6. 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
Courtesy Christy Johannesson

7. Zn (s) + 2 HCl (aq) ZnCl2(aq) + H2(g)
Gas Stoichiometry
Find vol. hydrogen gas made when 38.2 g zinc react w/excess hydrochloric acid.
Pres. = kPa; temp.= 88oC.
Zn (s) HCl (aq) ZnCl2(aq) H2(g)
38.2 g
excess
X L
P = kPa
(13.1 L)
T = 88oC
1 mol Zn
1 mol H2
22.4 L O2
x L H2 = g Zn
= L H2
65.4 g Zn
1 mol Zn
1 mol H2 Zn H2
The relationship between the amounts of gases (in moles) and their volumes (in liters) in the ideal gas law is used to calculate the stoichiometry of reactions involving gases, if the pressure and temperature are known.
• Relationship between the amounts of products and reactants in a chemical reaction can be expressed in units of moles or masses of pure substances, of volumes of solutions, or of volumes of gaseous substances.
• The ideal gas law can be used to calculate the volume of gaseous products or reactants as needed.
• In the lab, gases produced in a reaction are collected by the displacement of water from filled vessels — the amount of gas can be calculated from the volume of water displaced and the atmospheric pressure.
Combined
Gas Law
At STP, we’d use 22.4 L per 1 mol, but we aren’t at STP.
1 mol Zn
1 mol H2
x mol H2 = g Zn
= mol H2
65.4 g Zn
1 mol Zn
88oC = 361 K
V =
n R T P
0.584 mol (8.314 L.kPa/mol.K)(361 K)
P V = n R T = =
16.3 L
107.3 kPa

8. Zn (s) + 2 HCl (aq) ZnCl2(aq) + H2(g)
Gas Stoichiometry
Find vol. hydrogen gas made when 38.2 g zinc react w/excess hydrochloric acid.
Pres. = kPa; temp.= 88oC.
Zn (s) HCl (aq) ZnCl2(aq) H2(g)
38.2 g
excess
X L
P = kPa
(13.1 L)
T = 88oC
1 mol Zn
1 mol H2
22.4 L O2
x L H2 = g Zn
= L H2
65.4 g Zn
1 mol Zn
1 mol H2 Zn H2
The relationship between the amounts of gases (in moles) and their volumes (in liters) in the ideal gas law is used to calculate the stoichiometry of reactions involving gases, if the pressure and temperature are known.
• Relationship between the amounts of products and reactants in a chemical reaction can be expressed in units of moles or masses of pure substances, of volumes of solutions, or of volumes of gaseous substances.
• The ideal gas law can be used to calculate the volume of gaseous products or reactants as needed.
• In the lab, gases produced in a reaction are collected by the displacement of water from filled vessels — the amount of gas can be calculated from the volume of water displaced and the atmospheric pressure.
Combined
Gas Law
At STP, we’d use 22.4 L per 1 mol, but we aren’t at STP.
P1 =
T1 =
V1 =
P2 =
T2 =
V2 =
101.3 kPa
P1 x V1 T1
P2 x V2 T2
(101.3 kPa) x (13.1 L) = (107.3 kPa) x (V2)
273 K =
273 K
361 K
13.1 L
107.3 kPa V2 =
16.3 L
88 oC
+ 273
= 361 K
X L

9. 2 Mg (s) + CO2 (g) 2 MgO (s) + C (s)
What mass solid magnesium is required to react w/250 mL carbon dioxide
at 1.5 atm and 77oC to produce solid magnesium oxide and solid carbon? 2
Mg (s)
+ CO2 (g) 2
MgO (s)
+ C (s)
X g Mg
250 mL
0.25 L
V = 250 mL
0.25 L
oC = K
T = 77oC
350 K
P = 1.5 atm
kPa
n =
R T
P V
kPa
1.5 atm
(0.250 L)
P V = n R T
n =
= mol CO2
L.atm / mol.K
8.314 L.kPa / mol.K
(350 K)
2 mol Mg
24.3 g Mg
x g Mg = mol CO2
= g Mg
1 mol CO2
1 mol Mg
CO2 Mg

10. Gas Stoichiometry 2 Na + Cl2 NaCl 2 P1 x V1 T1 P2 x V2 T2 =
How many liters of chlorine gas are needed to react with excess sodium metal
to yield 5.0 g of sodium chloride when T = 25oC and P = 0.95 atm? 2
Na Cl NaCl 2
excess
X L
5 g
1 mol NaCl
1 mol Cl2
22.4 L Cl2
x g Cl2 = 5 g NaCl
= L Cl2
58.5 g NaCl
2 mol NaCl
1 mol Cl2
P1 x V1 T1
P2 x V2 T2
Ideal Gas
Method =
P1 = 1 atm
T1 = 273 K
V1 = L
P2 = atm
T2 = 25 oC = 298 K
V2 = X L
(1 atm) x (0.957 L) (0.95 atm) x (V2) =
273 K
298 K
V2 = L

11. Gas Stoichiometry 2 Na + Cl2 NaCl 2 V = n R T P P V = n R T V = 1.04 L
How many liters of chlorine gas are needed to react with excess sodium metal
to yield 5.0 g of sodium chloride when T = 25oC and P = 0.95 atm? 2
Na Cl NaCl 2
excess
X L
5 g
1 mol NaCl
1 mol Cl2
x g Cl2 = 5 g NaCl
= mol Cl2
58.5 g NaCl
2 mol NaCl
V =
n R T P
Ideal Gas
Method
P V = n R T
P = atm
T = 25 oC = 298 K
V = X L
R = L.atm / mol.K
n = mol
mol ( L.atm / mol.K) (298 K)
X L =
0.95 atm
V = L