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C3H8(g) + 2O2(g) → 4H2O(g) + 3CO2(g)
ΔHreaction=2,220 kJ/mol

Calculate the enthalpy of combustion for 24.5g propane given the molecular weight of propane is 44.097 g/mol.

A)
723 kJ


B)
1,010 kJ


C)
1,230 kJ


D)
2,390 kJ

Respuesta :

mnrochaar

Answer:

Option C) 1,230 kJ

Explanation:

According to the data, 1 mol of propane liberates 2,220 kJ of energy when it is combusted. If we have 24.5 g of propane, this mass is equal to 0,56 mol (of propane):

44.097 g of propane---->1 mol

24.5 g of propane-------x= 0.56 mol

Then, the enthalpy of the described chemical reaction (propane´s combustion) is 2,220 kJ when 1 mol of propane is combusted, so:

1 mol propane---------->2,220 kJ of energy

0.56 mol propane (24.5 g)------------x= aproximately 1,24 kJ (which would be option C)

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