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15 December, 00:28

What volume of benzene (C6H6, d = 0.88 g/mL, molar mass = 78.11 g/mol) is required to produce 1.5 x 103 kJ of heat according to the following reaction?2 C6H6 (l) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (g) ΔH°rxn = - 6278 kJ

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  1. 15 December, 02:31
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    We need 42.4 mL of benzene to produce 1.5 * 10³ kJ of heat

    Explanation:

    Step 1: Data given

    Density of benzene = 0.88 g/mL

    Molar mass of benzene = 78.11 g/mol

    Heat produced = 1.5 * 10³ kJ

    Step 2: The balanced equation

    2 C6H6 (l) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (g) ΔH°rxn = - 6278 kJ

    Step 3: Calculate moles of benzen

    1.5 * 10³ kJ * (2 mol C6H6 / 6278 kJ) = 0.478 mol C6H6

    Step 4: Calculate mass of benzene

    Mass benzene : moles benzene * molar mass benzene

    Mass benzene = 0.478 * 78.11 g

    Mass of benzene = 37.34 grams

    Step 5: Calculate volume of benzene

    Volume benzene = mass / density

    Volume benzene = 37.34 grams / 0.88g/mL

    Volume benzene = 42.4 mL

    We need 42.4 mL of benzene to produce 1.5 * 10³ kJ of heat
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