Why would "not flammable" be a chemical property rather than a physical property ?

1 answer:

4 0

When something is flammable, it burns and burning is a chemical property since the compounds are being broken down and gas is released. If something is non-falammable, it would then be a chemical property as well since the chemical makeup would not be broken by a flame. (I apologize if my answer is a bit broad or unspecific)

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The heat of combustion of propane, C3H8 (g) is -2057 kJ/mol. What would be the enthalpy change if enough propane was burned to g

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Considering the reaction stoichiometry, the enthalpy change if enough propane was burned to give off 12 moles of carbon dioxide gas is 8228 kJ.

The balanced reaction is:

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l)

The heat of combustion of propane, C₃H₈, is -2057 kJ/mol. This is, 2057 kJ is released for every 1 mol C₃H₈.

So to determine the enthalpy change if enough propane was burned to emit 12 moles of carbon dioxide, you must take into account the stoichiometry of the reaction.

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

C₃H₈: 1 mole
O₂: 5 moles
CO₂: 3 moles
H₂O: 4 moles

Then you can apply the following rule of three: if by stoichiometry 3 moles of CO₂ are produced by 1 mole of C₃H₈, 12 moles of CO₂ are produced by how many moles of C₃H₈?

$amount of moles of C_{3} H_{8} =\frac{12 moles of CO_{2}x1 mole of C_{3} H_{8} }{3 moles of CO_{2}}$

amount of moles of C₃H₈= 4 moles

So to determine the enthalpy change, you can apply the following rule of three: If for each mole of C₃H₈ 2057 kJ are released, for 4 moles of C₃H₈ how much heat is released?

$Heat released=\frac{4 molesx2057 kJ}{1 mole}$