The combustion of propane may be described by the chemical equation C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(g) How many grams of O2(g) are needed to completely burn 97.5 g C3H8(g)?

Question

The combustion of propane may be described by the chemical equation C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(g) How many grams of O2(g) are needed to completely burn 97.5 g C3H8(g)?

Answer 1

Hello, Joana,

It's nice to have a balanced equation to start. I checked, and it is correctly balanced. (Just in case . . .). In this reaction, the equation tells us we need 5 moles O₂ for every 1 mole C₃H₈. We can write this as a conversion factor:

(5 moles O₂)/(1 mole C₃H₈)

In other words, we'll consume 5 moles O₂ for every 1 mole C₃H₈.

Convert 97.5 grams of C₃H₈ to moles by dividing the mass by it's molar mass (44g/mole).

97.5g/44g/mole = 2.216 moles C₃H₈

To completely burn 2.216 moles C₃H₈, we'll need 5 times that number of moles of O₂, which is 11.08 moles O₂.

Convert moles into grams by multiplying by the molar mass of O₂.

11.08g)*(32g/mole)=355 g O₂ (3 sig figs)

Bob