Avogadro's Law
Volumes of gases
- In 1811 the Italian scientist Amedeo Avogadro developed a theory about the volume of gases
- Avogadro’s law (also called Avogadro’s hypothesis) enables the mole ratio of reacting gases to be determined from volumes of the gases
- Avogadro deduced that equal volumes of gases must contain the same number of molecules
- At room temperature and pressure(RTP) one mole of any gas has a volume of 24 dm3
- The units are normally written as dm3 mol-1(since it is ‘per mole’)
- The conditions of RTP are
- a temperature of 20 oC
- pressure of 1 atmosphere
Stoichiometric relationships
- The stoichiometry of a reaction and Avogadro’s Law can be used to deduce the exact volumes of gaseous reactants and products
- Remember that if the gas volumes are not in the same ratio as the coefficients then the amount of product is determined by the limiting reactant so it is essential to identify it first
Worked example
Example 1
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
Answer:
Step 2: We can see that 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, we can say 40 cm3 of NO will be produced
Step 4: There will be of 70-40 = 30 cm3 of NH3 left over
Therefore the total remaining volume will be 40 + 30 = 70 cm3 of gases
Worked example
Example 2
The complete combustion of propane gives carbon dioxide and water vapour as the products:
C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g)
Determine the volume of oxygen needed to react with 150 cm3 of propane and the total volume of the gaseous products.
Answer
Step 1: The balanced equation shows that 5 moles of oxygen are needed to completely react with 1 mole of propane
Step 2: Therefore the volume of oxygen needed would be = 5 moles x 150 cm3 = 750 cm3
Step 3: The total number of moles of gaseous products is = 3 + 4 = 7 moles
Step 4: The total volume of gaseous products would be = 7 moles x 150 cm3 = 1050 cm3
