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According to the following reactions, would the burning of 5.50 g of methane (CH4) or propane (C3H8) release more heat?

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) ΔΗ = -2043 kJ

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ΔΗ = -890. kJ

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mickymike92

Based in their molar enthalpies of combustion, the burning of 5.50 g of methane (CH4) would release more heat than the burning of 5.50 g propane (C3H8).

What is molar enthalpy of combustion?

The molar enthalpy of combustion is the amount of heat evolved when one mole of a substance is completely burnt in oxygen.

The molar enthalpy of combustion of propane is -2043 kJ/mol

The molar enthalpy of combustion of methane is -890 kJ/mol

Molar mass of propane = 44.0 g

Molar mass of methane = 16.0 g

Moles of propane in 5.50 g = 5.5/44 = 0.125 moles

Moles of methane in 5.50 g = 5.5/16 = 0.34375 moles

Heat released by 0.125 moles of propane = -2043 × 0.125 = -255.4 kJ

Heat released by 0.125 moles of methane = -890 × 0.34375 = -305.9 kJ

Therefore, the burning of 5.50 g of methane (CH4) would release more heat than the burning of 5.50 g propane (C3H8).

Learn more about molar enthalpy of combustion at: https://brainly.com/question/23937492

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