Module 5.04
Gas Stoichiometry
Stoichiometry Using the Ideal Gas Law
• If you are given the volume, pressure and
temperature of a gaseous reaction you can
use the ideal gas law to solve for moles
before solving a stoichiometry problem.
• n = PV
RT
n = # of moles
P = Pressure in atm
V = Volume in Liters
R = Ideal gas constant in 0.0821 L x atm
Mol x K
Stoichiometry
• It’s a process that allows us to
mathematically convert and calculate
the relationships between the amount of
reactants and products in a chemical
reaction.
Example 1 (Ideal gas law before stoichiometry)
• How many grams of sodium chloride can
react with 18.3 liters of fluorine gas at 1.2
atmospheres at 299 kelvins?
F2(g) + 2 NaCL(s)
Cl2 (g) + 2NaF (s)
1. Make sure your equation is balance
2. Always write down what you know
n=PxV
RxT
V = 18.3 L
P = 1.2 atm
T = 299 K
R = 0.0821 L atm
Mol K
Example 1 (con’t)
n=PxV
RxT
n(F2) = 1.2 atm x 18.3 L
0.0821 x 299K
n(F2)= 0.895 mol F2
Now that you have the amount of F2 in moles, you
can start the stoichiometry problem.
0.895 mol F2 x 2 mol NaCl x 58.44 g NaCl
= 105 g NaCl
1 mol F2
1 mol NaCl
Practice 1
How many grams of sodium metal are needed to react
completely with 17.8 liters of chlorine gas at 297 kelvins
and 1.25 atmospheres?
2Na(s) + Cl2(g)  2 NaCl(s)
Example 2 (Ideal gas law after stoichiometry)
If you are asked to solve for the volume or pressure of a gaseous
Product, you can first solve for moles of that product using stoichiometry
and then plug that value into the ideal gas law.
V = nRT or
P
P = nRT
V
Example 2
If 52.0 g of magnesium metal react with excess hydrochloric
acid, how many liters of hydrogen gas can be produced at 27 C
And 0.97 atmospheres?
Mg(s) + 2HCl(aq)  H2(g) + MgCl2 (aq)
1 mol Mg x 1 mol H2
52.0 g Mg x
24.3 g Mg 1 mol Mg
V=
(2.14 mol H2)(0.0821)(300 K)
0.97 atm
V (H2) =
54.3 L of H2 gas
= 2.14 mol H2
Practice 2
If 45.8 grams of lithium metal react with excess water, how
many liters of hydrogen gas will be produced at 27.5 C and
1.35 atmospheres?
2 Li (s) + 2 H2O (l)  2 NaOH (aq) + H2 (g)
Mole Ratios and Coefficients
2 CO (g) + O2 (g)  2 CO2 (g)
The coefficients in this balanced equation provide a volume
ration between gases when they are at the same
temperature and pressure.
A given volume of oxygen gas will react with twice that
volume of carbon dioxide, because the ration of O2 to CO is
1:2 (given by the coefficients)
The coefficients provide volume ratios (only gaseous reactants
and products) that can be used in stoichiometry calculations
involving volume.
Practice 1
2 CO (g) + O2 (g)  2 CO2 (g)
Assuming all volume measurements are made at the same
temperature and pressure, how many milliliters of carbon
dioxide gas be produced when 206.5 milliliters of oxygen gas
react with excess carbon monoxide?
Practice 2
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
How many liters of propane gas (C3H8) need to be combusted to
produce 3.8 liters of carbon dioxide, if all measurements are
taken at the same temperature and pressure?
Temperature and Pressure
STP = standard temperature and pressure
T = 273.15 kelvins
P = 1.0 atm
1 mol of gas
22.4 L
or
22.4 L
1 mol of gas
These ratios can be used within a stoichiometry calculation
whenever conditions are at STP
This relationship between volume and moles replaces the
need to use the ideal gas law.
Practice 1
How many grams of calcium oxide (lime) would be produced
when 50.0 liters of carbon dioxide is produced at STP?
CaCO3 (s)  CaO (s) + CO2 (g)
Diffusion and Effusion
Diffusion - The gradual mixing of two gases because of the
spontaneous, random motion of their particles.
Effusion - The movement of gas particles through a small
opening in the container wall due to the pressure and
particle movement inside the container.
The relationship between mass and velocity is expressed
by the gas law known as Graham’s Law:
√ Molar mass of B
Rate of effusion of A
=
Rate of effusion of B
√ Molar mass of A
Variations of Graham’s Law
rate of effusion of A
rate of effusion of B
=
√ Molar mass of B
√ Molar mass of A
√ Molar mass of B
velocity of A
=
velocity of B
√ Molar mass of A
rate of diffusion of A
rate of diffusion of B
=
√ Molar mass of B
√ Molar mass of A
Practice 1 (effusion)
A sample of hydrogen gas (H2) effuses through a porous
7.96 times faster than an unknown gas. What is the molar
mass of the unknown gas?
Practice 2 (diffusion)
Under certain conditions argon (Ar) gas diffuses at a rate of
5.2 centimeters per second. Under the same conditions, an
unknown gas diffuses at a rate of 3.5 centimeters per
second. What is the approximate molar mass of the unknown
gas?
Practice 3
Which sample of gas would effuse the fastest under the
same temperature and pressure conditions? Use the periodic
table to help answer this question.
CH4
Cl2
O2
Ne