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 32.2 g C3H8(g)?

Answer 1

Hello, Jaylin,

The balanced equation says we need 5 moles of O₂ for every 2 moles of C₃H₅. That is a molar ratio of:

(5 moles O₂)/(2 moles C₃H₅)

The 32.2 grams of propane should be coverted into moles propane. The molar mass of C₃H₅ is 44,03 grams/mole.

The moles of propane = (32.2 g).(44.03 g/mole) = 0.731 moles C₃H₅

We can convert this into the moles of O₂ needed by multiplying by the molar ratio.

(0.731 moles C₃H₅)*(5 moles O₂)/(2 moles C₃H₅) = 3.66 moles O₂

Now convert moles O₂ to grams O₂ by multiplying times the molar mass of O₂ (32 grams/mole):

(3.66 moles O₂)*(32 grams/mole O₂) = 117 grams O₂ to 3 sig figs

Bob