This problem is describing the combustion experiment where by the heat of combustion of 1.66 g of liquid C6H14 can be calculated via calorimetry. The calorimeter (C=11.2kJ/°C) contains 2750. grams of water with a temperature rise of 3.54 °C, it means we can calculate the total heat released by the combustion of this hydrocarbon via:
[tex]Q_{combustion}=-(Q_{calorimeter}+Q_{water})[/tex]
Because the heat of combustion is released to the water and the calorimeter. Next, we break down each heat and subsequently plug in the given values:
[tex]Q_{combustion}=-(11200\frac{J}{\°C}*3.54\°C +2750.g*4.184\frac{J}{g\°C}*3.54\°C )\\\\Q_{combustion}=-80,379J=-80.4kJ[/tex]
After that, we calculate the moles of C6H14 by considering the given mass and its molar mass (86.18 g/mol):
[tex]n=1.66g*\frac{1mol}{86.18g} =0.01926mol[/tex]
And finally, we divide the total heat of combistion by the moles to get the heat of combustion:
[tex]\Delta H_{comb}=\frac{-80.4kJ}{0.01926mol} =4,174\frac{kJ}{mol}[/tex]
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