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14 February, 20:29

Calculate the number of grams of Mg needed for this reaction to release enough energy to increase the temperature of 78 mL of water from 29 ∘C to 78 ∘C.

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  1. 14 February, 22:20
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    We need 1.1 grams of Mg

    Explanation:

    Step 1: Data given

    Volume of water = 78 mL

    Initial temperature = 29 °C

    Final temperature = 78 °C

    The standard heats of formation

    -285.8 kJ/mol H2O (l)

    -924.54 kJ/mol Mg (OH) 2 (s)

    Step 2: The equation

    The heat is produced by the following reaction:

    Mg (s) + 2H2O (l) →Mg (OH) 2 (s) + H2 (g)

    Step 3: Calculate the mass of Mg needed

    Using the standard heats of formation:

    -285.8 kJ/mol H2O (l)

    -924.54 kJ/mol Mg (OH) 2 (s)

    Mg (s) + 2 H2O (l) → Mg (OH) 2 (s) + H2 (g)

    -924.54 kJ - (2 * - 285.8 kJ) = - 352.94 kJ/mol Mg

    (4.184 J/g·°C) * (78 g) * (78 - 29) °C = 15991.248 J required

    (15991.248 J) / (352940 J/mol Mg) * (24.3 g Mg/mol) = 1.1 g Mg

    We need 1.1 grams of Mg
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