Asked by Hunter
The thermochemical equation for the combustion of hexane is shown below.
2 C6H14 (g) + 19 O2 (g) → 12 CO2 (g) + 14 H2O (g) ΔH° = –8326 kJ
What is the enthalpy change for the combustion of 0.650 L C6H14 at STP?
2 C6H14 (g) + 19 O2 (g) → 12 CO2 (g) + 14 H2O (g) ΔH° = –8326 kJ
What is the enthalpy change for the combustion of 0.650 L C6H14 at STP?
Answers
Answered by
DrBob222
8326 kJ is produced by 2*22.4L C6H14. So for 0.650 L we will produce
8326 kJ x (0.650L/2*22.4L) = ? kJ.
8326 kJ x (0.650L/2*22.4L) = ? kJ.
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