Avogadro's Law (DP IB Chemistry): Revision Note
Avogadro's Law
Volumes of gases
In 1811, Amedeo Avogadro proposed that equal volumes of gases (under the same conditions of temperature and pressure) contain the same number of molecules
This is known as Avogadro’s Law, and it allows us to determine mole ratios from the volumes of reacting gases
At standard temperature and pressure (STP):
One mole of any gas occupies 22.7 dm3
Units are written as dm3 mol-1
STP conditions are defined as:
0 °C (273 K)
100 kPa pressure
Stoichiometric relationships
The volume ratio of gaseous reactants and products in a balanced chemical equation is the same as the mole ratio
You can use this ratio to calculate unknown gas volumes in a reaction at STP
Example: Combustion of propane
In the reaction:
C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (l)
If 50 cm³ of propane is burned:
Volume of O2 needed = 5 × 50 = 250 cm3
Volume of CO2 formed = 3 × 50 = 150 cm3
The mole ratio from the equation (1:5:3) is directly applied to gas volumes
Examiner Tips and Tricks
These gas volume relationships are only valid if all gases are measured under the same conditions of temperature and pressure
If gases are not in the same ratio as the balanced equation, use limiting reactant principles to determine the actual amount of product formed
Worked Example
What is the total volume of gases remaining when 70 cm3 of ammonia is combusted completely with 50 cm3 of oxygen according to the equation shown?
4NH3 (g) + 5O2 (g) → 4NO (g) + 6H2O (l)
Answer:
Step 1
From the equation deduce the molar ratio of the gases (water is in the liquid state so is not included
NH3 :O2 :NO or 4:5:4
Step 2
Oxygen will run out first (the limiting reactant) and so 50 cm3 of O2 requires 4/5 x 50 cm3 of NH3 to react = 40 cm3
Step 3
Using Avogadro's Law, 40 cm3 of NO will be produced
Step 4
There will be of 70-40 = 30 cm3 of NH3 left over
Total remaining = 40 + 30 = 70 cm3 of gases
Examiner Tips and Tricks
Since gas volumes work in the same way as moles, we can use the 'lowest is limiting' technique in limiting reactant problems involving gas volumes. This can be handy if you are unable to spot which gas reactant is going to run out first. Divide the volumes of the gases by the coefficients and whichever gives the lowest number is the limiting reactant
E.g. in the previous problem we can see that
For NH3
gives 17.5For O2
gives 10, so oxygen is limiting
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