Transcript Stoich with Gases!

Stoich with Gases!
How do we figure out how many
particles there are in a gas sample?
Avagadro’s Law
• Equal volumes of gases at the same
temperature and pressure contain equal
numbers of molecules
• The volume occupied by one mole of a gas at
STP is known as the standard molar volume of
a gas and is equal to 22.4 L.
Example Problem 1:
• A chemical reaction produces 0.0680 mol of
Oxygen gas. What volume in liters is occupied
by this gas sample at STP?
0.0680 mol x 22.4 L = 1.52 L O2
1 mol
Example Problem 2:
• A chemical reaction produced 98.0 mL of
sulfur dioxide gas, SO2, at STP. What was the
mass in grams of the gas produced?
98.0 mL x 1 L x 1 mol x 64.07 g SO2 = 0.28 g SO2
1000 mL 22.4 L 1 mol SO2
Your Turn
1. What is the mass of 1.33 x 104 mL of oxygen
gas at STP?
2. What is the volume of 77.0 g of nitrogen
dioxide gas at STP?
3. At STP, 3 L of chlorine is produced during a
chemical reaction. What is the mass of this
gas?
What is the mass of 1.33 x 104 mL of
oxygen gas at STP?
1.33 x 104 mL x 1 L x 1 mole x 32 g = 19 grams
1000 mL 22.4 L 1 mol
What is the volume of 77.0 g of
nitrogen dioxide gas at STP?
77.0 g NO2 x 1 mol NO2 x 22.4 L = 37.5 L
46 g NO2
1 mol
At STP, 3 L of chlorine is produced
during a chemical reaction. What is
the mass of this gas?
3 L Cl2 x 1 mole x 70.9 g = 9.5 grams
22.4 L
1 mol
What would happen to the pressure of
a gas in a fixed container if you added
more molecules of gas?
What would happen to the volume of
a gas in a flexible container if you
added more molecules of gas?
Volume, Temperature, Pressure, and
number of moles/molecules are all
related.
Ideal Gas Law: mathematical relationship
between pressure, volume, temperature, and
the number of moles of a gas
V = nRT
P
OR PV = nRT
Moles
Ideal Gas
Constant
PV = nRT
Ideal Gas Constant:
R = 0.0821 L·atm
mol·K
R = 8.314
J
mol·K
Example Problem:
• What is the pressure in atmospheres exerted
by a 0.500 mol sample of nitrogen gas in a
10.0 L container at 25°C?
25 + 273 = 298 K
PV = nRT
so P = nRT
V
P = (0.5 mol)(0.0821L·atm)(298 K) = 1.22 atm
10.0 L mol·K
Your Turn
• What is the volume in liters of 0.25 mol of
oxygen gas at 20°C and 0.974 atm pressure?
20 + 273 = 293 K
PV = nRT
so V = nRT
P
V = (0.25 mol)(0.0821L·atm)(293 K) = 6.17 L O2
0.974 atm
mol·K
Another
• What mass of chlorine gas, Cl2, in grams, is
contained in a 10.0 L tank at 27°C and 3.50
atm of pressure?
27 + 273 = 300 K
PV = nRT
so n = PV
RT
n = (3.50 atm)(10.0 L) = 1.42 mol Cl2 x 70.9 g = 101 g Cl2
(0.0821L·atm)(300 K)
1 mol
mol·K
Stoichiometry of Gases
• The coefficients in chemical equation not only
indicate mole ratios, but also reveal volume
ratios.
• 2CO(g)
+
O2(g) 
2 CO2 (g)
2 molecules
1 molecule
2 molecule
2 mol
1 mol
2 mol
2 volumes
1 volume
2 volumes
Stoichiometry of Gases
• Propane C3H8 is used for cooking and heating.
The combustion of propane occurs according
to the following equation:
C3H8(g) + 5O2(g)  3 CO2 (g) + 4 H2O(g)
a) What is the volume, in liters, of oxygen
required for the complete combustion of 0.350L
of propane?
b) What will be the volume of carbon dioxide
produced?
Putting it all together: Calcium carbonate
(limestone) can be heated to produce calcium oxide
(lime). The balanced equation is
CaCO3(s)  CaO(s) + CO2(g)
How many grams of CaCO3 must be decomposed to
produce 5.00 L of carbon dioxide gas at STP?
STP: T = 273. K P = 1.00 atm
PV = nRT
so n = PV
RT
n = (1.00 atm) (5.00 L) = 0.223 mol CO2x 1 mol CaCO3 x 100.08 g CaCO3 = 22.3 g
(0.0821L·atm)(273 K)
1 mol CO2
1 mol CaCO3
mol·K
First use ideal gas law to calculate
the number of moles of CO2
produced
Then use mole ratio and molar
mass to calculate the mass of
CaCO3 needed.
Putting it all together: Calcium carbonate
(limestone) can be heated to produce calcium oxide
(lime). The balanced equation is
CaCO3(s)  CaO(s) + CO2(g)
How many grams of CaCO3 must be decomposed to
produce 5.00 L of carbon dioxide gas at STP?
STP: T = 273 K P = 1.00 atm
Method 2: at STP 1.00 mol of gas = 22.4 L of gas
5.00 L CO2 x 1.00mol CO2 x 1 mol CaCO3 x 100.08 g CaCO3 = 22.3 g
22.4 L CO2
1 mol CO2
1 mol CaCO3
One more: Tungsten, W, is produced industrially by
the reaction of tungsten oxide with hydrogen
WO3(s) + 3H2(g)  W(s) + 3H2O(g)
How many liters of hydrogen gas at 35°C and 0.980
atm are needed to react completely with 875 g of
tungsten oxide?
875g WO3 x 1mol WO3 x 3mol H2 = 11.3mol H2
231.84g WO3 1 mol WO3
First find the number of moles using stoichiometry, then use ideal gas law to find V.
T =35+ 273= 308 K P = 0.980 atm n = 11.3mol H2
PV = nRT
𝐿 𝑎𝑡𝑚
(11.3molH2)(0.0821
)(308K)
𝑚𝑜𝑙 𝐾
so V = nRT =
P
= 292 L H2
0.980 atm