Taking into account definition of STP conditions and reaction stoichiometry, 1384.992 L of oxygen are necessary for the combustion of 277L of propane at STP.
In first place, the balanced reaction is:
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
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:
The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.
You can apply the following rule of three: if by definition of STP 22.4 L of propane is occupied by 1 mole, 277 L is occupied by how many moles of propane?
[tex]moles of propane= \frac{277 Lx1 mole}{22.4 L}[/tex]
moles of propane= 12.366 moles
You can apply the following rule of three: If by reaction stoichiometry 1 mole of propane react with 5 moles of O₂, 12.366 moles of propane react with how many moles of O₂?
[tex]moles of O_{2} =\frac{12.366 moles of propanex5 moles of O_{2} }{1 moles of propane}[/tex]
moles of O₂= 61.83 moles
Finally, you can apply the following rule of three: if by definition of STP conditions 1 mole of O₂ occupies a volume of 22.4 liters, 61.83 moles occupies how much volume?
[tex]volume=\frac{61.83 molesx22.4 L}{1 mole}[/tex]
volume= 1384.992 L
In summary, 1384.992 L of oxygen are necessary for the combustion of 277L of propane at STP.
Learn more about
the reaction stoichiometry:
brainly.com/question/24741074
brainly.com/question/24653699
STP conditions:
https://brainly.com/question/1186356
https://brainly.com/question/8862004
