21. Calculate the volume of carbon dioxide at 20.0 degrees C and 0.910 atm produ

Question

21. Calculate the volume of carbon dioxide at 20.0 degrees C and 0.910 atm produced from the complete combustion of 6.50 kg of methane. Compare your result with the volume of CO2 produced from the complete combustion of 6.50 kg of propane (C3H8). a. Volume of CO2 from 6.50 kg mathane? answer in L b. Volume of CO2 from 6.50 kg propane? answer in L

Explanation / Answer

FOLLOW THIS Moles of CO2: CH4 + 2O2 -------> CO2 + 2H2O so: 1 mole of CH4 produces 1 mole of CO2 - i.e. 16 g of CH4 will produce 44 g of CO2 when it reacts with oxygen You used 2000g CH4. Use proportions: 16/44 = 2000/CO2 produced from 2000g, then 16 CO2 = 44 X 2000 ; CO2 = 5500 g. 5500 g divided by the molecular wight of CO2 = 5500/44 = 125 moles Another way is to look at the methane. 2000g/16 grams/mole = 125 moles. The equation says that 1 mole methane will make 1 mole CO2, so 125 moles will make 125 moles CO2 The propane reaction can be handled similarly, but the mole proportions will differ: C3H8 + 5O2 -----> 3CO2 + 4H2O That is 1 mole of propane makes 3 moles of CO2 - proceed from there. I do hope you have some input (from a teacher or text) on what the specific equations are - you are lost without them as I would be.

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