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 80.5 g C3H8(g)?

Answer 1

There is a 1:5 ratio of mol propane to mol O₂, which means that for every mol of propane, 5 mol O₂ is required to combust it.

Let's first find the mol propane:

1. molar mass of propane = 3(molar mass of carbon) + 8(molar mass of hydrogen) = 3(12.011) + 8(1.008) = 44.097 g/mol propane
2. 80.5g propane ÷ 44.097 g/mol propane = 1.826 mol propane

In order to combust 1.826 mol propane, we need 5 times the amount of O₂ --> 5(1.826) = 9.13 mol O₂

To find the grams O₂, we multiply by the molar mass, which is 2(molar mass of oxygen) = 2(16) = 32

9.13(32) = 292.16g O₂

Without rounding in between, a more accurate result would be 292.083g O₂ required to combust 80.5g propane.