White gas is a mixture of C5 compounds and C6 compounds. The heavier hydrocarbon
ID: 834114 • Letter: W
Question
White gas is a mixture of C5 compounds and C6 compounds. The heavier hydrocarbons in white gas are hexanes (C6H14).
1. How much heat is release during the combustion of 3.00 kg of C6H14?
2.How many grams of C6H14 are needed to heat 1.25 kg of water from 23.5C to 80.5C ? Assume that all of the heat released during combustion is use to heat the water.
3. Assume white gas is 25%C5 hydrocarbons (fuel value=48.99 kj/g) and 75%C6 hydrocarbons. how many grams of white gas are needed to heat 1.25 kg of water from 23.5C to 80.5C?
Explanation / Answer
delta H combustion of hexanes (C6H14) = - 4163 kJ/mol
the molar mass of C6H14 is 86.18 g/mol
dH = ( - 4163 kJ) / (86.18 grams) = - 48.306 kJ / gram
dH = - 48.31 kJ / gram
1. (3000 grams ) ( - 48.31 kJ / gram) = 144,930 kJ are released per kg C6H14
2. dH = m C dT
dH = (1.25 kg) (4.184 kJ/kg-C) (57 Celsius rise in temp)
dH = 298.11 kJ required
grams of C6H14 are needed
(298.11 kJ required) / ( 48.31 kJ / gram) = 6.17 grams of C6H14 are required
3. (25% of 48.99 kJ/g) + (75% of 48.31 kJ / gram) = total
12.2475 + 36.2325 = 48.48 kJ/gram for the mixture
find kJ needed to heat 1.25kg of water from 23.5 degrees celsius to 80.5 degrees celsius
dH = m C dT
dH = (1.25 kg) (4.184 kJ/kg-C) (57 Celsius rise in temp)
dH = 298.11 kJ required
grams of white gas are needed
(298.11 kJ required) / (48.48 kJ/gram) = 6.14 grams of mix is required
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