For the combustion of propane (C3H8), how many grams of oxygen are needed to fully react 45.0 grams of propane?

Answer 1

Hello, Alture,

I'm not clear as to what portion of this question you'd like help on. I'll provide an overview of the process, but please respond as to which part is most confusing and I'll explain it more thoroughly if you'd like.

1) We need a balanced equation to start. I came up with:

C₃H₈ + 5O₂ = 3CO₂ + 4H₂O

We can see that the molar ratio of O₂ to C₃H₈ is 5 (moles O₂/1 mole C₃H₈ ). We need 5 moles of Ooxygen for every one mole of C₃H₈ .

2) Calculate moles C₃H₈ provided: (45.0 grams C₃H₈ )/(44.03 g/mole) = 1.022 moles C₃H₈

3) Calculate moles O₂ required to combust all the C₃H₈ :

(1.022 moles C₃H₈ )*(5 moles O₂/1 mole C₃H₈) = 5.11 moles O₂

4) Convert moles oxygen into grams oxygen:

(5.11 moles O₂)*(32 grams/mole) = 163 grams O₂ to 3 sig figs

Bob