Question

Consider the following reaction representing the combustion of propane (C₃H₈): C₃H₈+ O₂ = CO₂ + H₂ If air contains 21 percent oxygen (v/v), what volume of air at STP would be required to burn 240 g of propane?

Answer 1

C₃H₈ + 5 O₂ --------------> 3 CO₂ + 4 H₂O

mass of propane = 240 g

moles of propane = 240 / 44.0956 = 5.44 mol

1 mol C3H8   ---------> 5 mol O2

5.44 mol C3H8   ---------> ??

moles of O2 = 5.44 x 5 = 27.2 mol

P V = n R T

1 x V = 27.2 x 0.0821 x 273

V = 609.9 L

volume of O2 = 609.9 L

volume of Air = 609.9 / 0.21

volume of Air = 2904 L