Asked by Riese
When the equation for combustion for ethane is balanced using integer coefficients, the ΔH for the reaction = -2834 kJ. How many grams of ethane must be burned in order to heat 277.3 grams of water from 54.0°C to the boiling point and then boil all of it into the gas phase at 100.0°C? The ΔHvap of water = 40.8 kJ/mol and the specific heat of liquid water (SH2O) = 4.184 J/gK.
Answers
Answered by
DrBob222
2C2H6 + 7O2 --> 4CO2 + 6H2O + 2834 kJ
How many kJ does it take to do the water thing.
q1 = heat needed to heat H2O from 54 C to 100.
q1 = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q2 = heat needed to boil away the water.
q2 = mass H2O x heat vaporiation.
Total heat needed is q1 + q2 and change to kJ.
2*molar mass C2H6 x (kJ needed/2834 kJ) = grams C2H6 needed.
Post your work if you get stuck.
How many kJ does it take to do the water thing.
q1 = heat needed to heat H2O from 54 C to 100.
q1 = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q2 = heat needed to boil away the water.
q2 = mass H2O x heat vaporiation.
Total heat needed is q1 + q2 and change to kJ.
2*molar mass C2H6 x (kJ needed/2834 kJ) = grams C2H6 needed.
Post your work if you get stuck.
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