Taking into account the reaction stoichiometry, 4.367×10²³ molecules of oxygen are formed when 58.6 g of KNO₃ decomposes.
Reaction stoichiometry
In first place, the balanced reaction is:
4 KNO₃ → 2 K₂O + 2 N₂ + 5 O₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- KNO₃: 4 moles
- K₂O: 2 moles
- N₂: 2 moles
- O₂: 5 moles
Being the molar mass of KNO3 101.11 g/mol, if 58.6 g of KNO₃ decomposes, the moles of the compound that react are calculated are:
moles of KNO₃= 58.6 g÷ 101.11 g/mole
moles of KNO₃= 0.58 moles
Molecules of oxygen formed
The following rules of three can be applied: if by reaction stoichiometry 4 moles of KNO₃ form 5 moles of O₂, 0.58 moles of KNO₃ form how many moles of O₂?
moles of O₂= (0.58 moles of KNO₃×5 moles of O₂)÷ 4 moles of KNO₃
moles of O₂= 0.725 moles
Avogadro's Number is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023×10²³ particles per mole. Avogadro's number applies to any substance.
Then you can apply the following rule of three: If 1 mole of oxygen contains 6.023×10²³ molecules, 0.725 moles of oxygen contains how many molecules?
amount of molecules= (6.023×10²³ molecules× 0.725 moles)÷ 1 mole
amount of molecules= 4.367×10²³ molecules
Finally, 4.367×10²³ molecules of oxygen are formed when 58.6 g of KNO₃ decomposes.