contestada

How many moles of oxygen gas would be needed to react with 155 g of propane gas,
C3Hg, in a combustion reaction?
C3H8 (g) + 502 (g)
+
-> 3CO2 (g) + 4H20 (1)

Respuesta :

CintiaSalazar

Taking into account the reaction stoichiometry, 17.61 moles of O₂ is required to react with 155 grams of C₃H₈.

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

The molar mass of the compounds is:

  • C₃H₈: 44 g/mole
  • O₂: 32 g/mole
  • CO₂: 44 g/mole
  • H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • C₃H₈: 1 mole ×44 g/mole= 44 grams
  • O₂: 5 moles ×32 g/mole= 160 grams
  • CO₂: 3 moles ×44 g/mole= 132 grams
  • H₂O: 4 moles ×18 g/mole= 72 grams

Mass of O₂ required

The following rule of three can be applied: if by reaction stoichiometry 44 grams of C₃H₈ react with 5 moles of O₂, 155 grams of C₃H₈ react with how many moles of O₂?

[tex]moles of O_{2} =\frac{155 grams of C_{3} H_{8}x5 moles of O_{2} }{44 grams of C_{3} H_{8}}[/tex]

moles of O₂= 17.61 moles

Finally, 17.61 moles of O₂ is required to react with 155 grams of C₃H₈.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

Otras preguntas