Answered

Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

How many liters of oxygen are necessary for the

combustion of 277L of propane (C,H,) at STP?

C3H8 +502–>3 CO2+ 4 H2O

Sagot :

CintiaSalazar

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.

Reaction stoichiometry

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:

  • C₃H₈:  1 mole  
  • O₂: 5 moles
  • CO₂: 3 moles
  • H₂O: 4 moles

STP conditions

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.

Moles of propane reacted

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

Moles of oxygen necessary

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

Volume of oxygen necessary

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

Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.