Which statement best explains why heating a liquid affects its viscosity?
2 answers:
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The heat of reaction (i.e. combustion) of butane (
) when reacted with oxygen (
) is -2658 kJ/mol butane, and the chemical reaction is given by:
+ 
---> 
+ 
The mass of butane required in the reaction is based on the heat produced by the reaction, which is given to be -1,500 kJ. The minus sign is added because the reaction releases heat (exothermic), which means that the products are in a "lower energy state" than the reactants.
Dividing this with the heat of reaction per mole of butane reacted would give the number of moles butane required. Then, multiplying the answer with the molar mass of butane which is 58 grams/mole, will give the mass of butane required.
Moles of butane = [(-1,500 kJ)/(-2658 kJ/mol butane)]
Moles of butane = 0.5643 moles butane
Mass of butane = 0.5643 moles butane * 58 grams/mol butane
Mass of butane = 32.73 grams butane
The mass of carbon dioxide (
) can be determined by multiplying the moles of butane () with the mole ratio of () produced to the () reacted, and then with the molar mass of (), which is 44 grams/mole.
Mass of carbon dioxide produced
= 0.5643 moles butane * [4 moles/ 1 mole ] * 44 grams/mole
Mass of carbon dioxide produced
= 99.32 grams
Thus, the mass of butane required is 32.73 grams, and the mass of carbon dioxide produced from the reaction of this amount of butane is 99.32 grams.
The mass of butane required in the reaction is based on the heat produced by the reaction, which is given to be -1,500 kJ. The minus sign is added because the reaction releases heat (exothermic), which means that the products are in a "lower energy state" than the reactants.
Dividing this with the heat of reaction per mole of butane reacted would give the number of moles butane required. Then, multiplying the answer with the molar mass of butane which is 58 grams/mole, will give the mass of butane required.
Moles of butane = [(-1,500 kJ)/(-2658 kJ/mol butane)]
Moles of butane = 0.5643 moles butane
Mass of butane = 0.5643 moles butane * 58 grams/mol butane
Mass of butane = 32.73 grams butane
The mass of carbon dioxide (
Mass of carbon dioxide produced
= 0.5643 moles butane * [4 moles
Mass of carbon dioxide produced
= 99.32 grams
Thus, the mass of butane required is 32.73 grams, and the mass of carbon dioxide produced from the reaction of this amount of butane is 99.32 grams.