Ethanol (C 2 H 5 OH) is currently blended withgasoline as an automobile fuel. (a

Question

Ethanol (C2H5OH) is currently blended withgasoline as an automobile fuel. (a) Write abalanced equation for the combustion of liquid ethanol in air.(b) Calculate the standard enthalpy change for thereaction, assuming H2O(g) as a product.(c) Calculate the heat produced per liter ofethanol by combustion of ethanol under constant pressure. Ethanolhas a density of 0.789 g/mL. (d) Calculate themass of CO2 produced per kJ of heat emitted.

I was able to figure out the answers to a and b, I just am not surehow to do c and d.
The answers I got for a and b are:
(a) C2H5OH(l) +3O2(g) ---> 2CO2(g) +3H2O(g).
(b) (2(-393.5) + 3(-241.82)) - ((-277.7) + 3(0)) =-1234.8 kJ

Any help will be greatly appreciated!

Explanation / Answer

(c) C2H5OH(l) + 3O2(g) --->2CO2(g) + 3H2O(g) : H = - 1234.8KJ Density of ethanol , d = 0.789 g / mL That means 1 mL of ethanol weighs 0.789 g So , 1L = 1000 ml of ehtanol weighs 0.789 * 1000 = 789 g Molar mass of C2H5OH is = 2 * 12 + 5 * 1 + 16 + 1 = 46 g 46 g of ethanol on complete combustion the heat produced is1234.8 KJ 789 g of ethanol on complete combustion the heat producedis X KJ X =( 1234.8 * 789 ) / 46     = 21179.5 KJ Molar mass of CO2 = 12 + 2 * 16 = 44 g For the production of 2 * 44 g of CO2 the heat produced is1234.8 KJ For the evolution of 1 KJ of heat the mass of CO2 produced is = ( 2 * 44 ) / 1234.8                                                                                                 = 0.0712 g                                                                                                  

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