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some camping stoves contain liquid C4H10 they work only when the outside temperature is warm enough
the butane to have a reasonable vapor pressure so they are not very good for camping in temperature below about 0 centigrade. Assume the enthalpy of vaporization of butane is 24.3 kj/mol . if the camp stove fuel tank contains 190.g liquid c4h10 how much heat energy transfer is required to vaporize all of the butane?
1 Answers
First you must assume that all of the Butane is liquid in the pressurized fuel tank (the amount of gaseous Butane is very small in comparison).
The 190. grams of Butane / 58.124 g/mol = 3.269 moles of Butane.
Multiply this number of moles by 24.3 kJ/mol to get the total heat required to vaporize all of the butane.
The 190. grams of Butane / 58.124 g/mol = 3.269 moles of Butane.
Multiply this number of moles by 24.3 kJ/mol to get the total heat required to vaporize all of the butane.
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