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3 December, 17:37

2C3H7OH (l) + 9O2 (g) → 8H2O (g) + 6CO2 (g)

ΔHreaction = 1,830 kJ/mol

Calculate the enthalpy of combustion for 55.9g isopropanol given the molecular weight of isopropanol is 60.096 g/mol.

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  1. 3 December, 18:05
    0
    850.95 kj

    Explanation:

    Given dа ta:

    ΔH of reaction = 1830 kj/mol

    Enthalpy of combustion for 55.9 g of isopropanol = ?

    Solution:

    Number of moles of isopropanol:

    Number of moles = mass / molar mass

    Number of moles = 55.9 g / 60.096 g/mol

    Number of moles = 0.93 mol

    1830 kj heat is produced when 2 mole of isopropanol react.

    For one mole:

    1830 kj/2 = 915 kj/mol

    For 0.93 mol

    0.93 mol * 915 kj/mol = 850.95 kj
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