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The combustion of propane is given by the following reaction. C 3 H 8 + 3 O 2 → 3 CO 2 + 4 H 2 O The enthalpy of reaction is −2202.0 kJ/mol. How much energy (in joules) will be released if 72.82 grams of propane is burned. (Molar mass of propane = 44.11 g/mol). Record your answer in scientific notation using 3 significant figures.

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Answer:

3635kJ of will be released

Explanation:

Based in combustion of propane:

C3H8 + 5O2 → 3CO2 + 4 H2O + ΔH

Where ΔH is the heat released per mole of propane that reacts (This heat is -2202.0kJ/mol).

Thus, we need to find the moles of propane that reacts using molar mass of propane to find the released heat as follows:

Moles propane:

72.82g * (1mol / 44.11mol) = 1.651 moles of propane will react

As 1 mole release 2202.0 kJ, 1.651 moles will release:

1.651 moles propane * (2202.0 kJ/mol) =

3635kJ of will be released

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