Calculate the enthalpy change (ΔH) for the combustion of 3 moles of methane (CH4) to produce carbon dioxide (CO2) and water (H2O), given the heat of combustion (ΔHcomb) of methane is -890.4 kJ/mol.
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The enthalpy change (ΔH) for the combustion of 3 moles of methane can be calculated using the stoichiometry of the reaction and the given heat of combustion (ΔHcomb) of methane.
The balanced equation for the combustion of methane is:
CH4 + 2O2 → CO2 + 2H2O
From the balanced equation, we can see that 1 mole of methane produces 1 mole of carbon dioxide and 2 moles of water.
Therefore, for the combustion of 3 moles of methane, we will have:
3 moles CH4 + 3 * (2 moles O2) → 3 moles CO2 + 3 * (2 moles H2O)
From the reaction stoichiometry, we can see that the molar ratio of methane to carbon dioxide is 1:1.
Therefore, the enthalpy change (ΔH) for the combustion of 3 moles of methane to produce carbon dioxide and water can be calculated as:
ΔH = ΔHcomb * moles of methane
= -890.4 kJ/mol * 3 moles
= -2671.2 kJ/mol
Therefore, the enthalpy change (ΔH) for the combustion of 3 moles of methane is -2671.2 kJ/mol.
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