how much oxygen is consumed in equation: C5H12 + 8O2 → 6H2O + 5CO2 when 10.0 grams of the pentane is burned in excess oxygen.

Question

please show your work, this is the reaction when 10.0 grams of propane are burned in 10.0 grams of oxygen

Answer 1

Complete combustion of pentane: C5H12 + 8O2 → 6H2O + 5CO2 ... balanced equation

moles C5H12 used = 10.0 g C5H12 x 1 mol/72 g = 0.139 moles pentane

moles O2 consumed = 0.139 mol C5H12 x 8 mole O2/1 mol C5H12 = 1.11 moles O2

mass O2 consumed = 1.11 moles O2 x 32 g/mole = 35.5 g O2

Combustion of propane: C3H8 + 5O2 ==> 3CO2 + 4H2O ... balanced equation

moles C3H8 = 10.0 g x 1 mol C3H8/44 g = 0.227 moles

moles O2 = 10.0 g O2 x 1 mol O2/32 g = -.313 moles

Since it takes 5 mol O2 for every mole of C3H8, O2 will be limiting. This complicates the problem slightly because you will not have complete combustion of propane.

Now, what is the question? Are you asking what is limiting? Are you asking how much CO2 or H2O is formed? You didn't complete the question.