Answer:- -2260 kj.
Solution:- It could easily be solved using dimensional analysis. From given information, the enthalpy change for the combustion of 2.00 grams of propane is -102.5 kj. It asks to calculate the enthalpy change in same conditions for the combustion of 1 mole of propane.
We convert 1 mole of propane to grams using molar mass(44.1 gram per mol) and then calculate the heat using the given information as:
[tex]1mol(\frac{44.1g}{mol})(\frac{-102.5kj}{2.00g})[/tex]
= -2260 kj
Hence, the enthalpy change for the combustion of 1 mol of propane is -2260 kj.
FURTHER EXPLANATION
The molar enthalpy of combustion is the amount of heat released by burning one mole of the substance. It can be determined from the heat of combustion obtained through calorimetry. The following conversion factors may be used:
2.00 g C₃H₈ = -102.5 kJ
1 mol C₃H₈ = 44.097 g
To solve the molar enthalpy of combustion, the following dimensional analysis must be setup:
[tex]molar \ enthalpy \ of \ combustion = 1 \ mol \ C_3H_8 \times \frac{44.097 \ g}{1 \ mol} \times \frac{-102.5 \ kJ}{2.00 \ g}\\\\\boxed {molar \ enthalpy \ of \ combustion \ = -2259.971 kJ}[/tex]
The least number of significant figures in the given is 3 (2.00 g) so the final answer must also have the 3 significant figures. Hence,
[tex]\boxed {\boxed {molar \ enthalpy \ of \ combustion \ = -2260 kJ}}[/tex]
This answer is negative because combustion reactions are exothermic. The magnitude of the molar enthalpy is greater than the sample's (given) because the propane used in calorimetry is much less than 1 mole. Therefore, the final answer is logical.
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Keywords: enthalpy, calorimetry, heat of combustion
