Propane (C3 H8) reacts with oxygen (O2) to produce a carbon dioxide and water. I

Question

Propane (C3 H8) reacts with oxygen (O2) to produce a carbon dioxide and water. In a particular experiment 38.0 grams of carbon dioxide are produced from the reaction of 22.05 grams of propane with excess oxygen. What is the % yield in this reaction?

Explanation / Answer

C3H8 + 5O2 --->3 CO2 + 4H2O , 38 gm CO2 = 38/44 =0.8636 moles CO2 , 22.05 gm propane = (22.05/44.1) = 0.5 moles , expected CO2 moles = 3 x0.5= 1.5 moles as per stochiometry in equation , CO2 in gm expected = 1.5 x44 = 66 gm , % yield = ( observed mass/ expected mass) x100 =( 38/66) x100 = 57.57 %

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