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13 August, 22:53

Calculate the work done by the expansion of hydrogen gas when 7.1 g of sodium react with water at constant temperature of 27°C and pressure of 1.7 atm. Use a precision of 2 significant digits. 2Na (s) + 2H2O (l) → 2NaOH (s) + H2 (g)

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  1. 14 August, 00:50
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    -390J

    Explanation:

    The given reaction is,

    2Na (s) + 2H2O - -> 2NaOH + H2

    Number of moles of NaOH = Mass/molar mass

    = 7.1/23 = 0.31 mol

    Since, 2 mol Na produces 1 mol H2

    Hence, 0.31 mol Na will produce

    = 0.31 mol / 2 = 0.155 mol H2

    According to ideal gas equation,

    PV = n RT

    V = nRT/P

    = 0.155mol x 0.0821 L atm K-1 mol-1 x (27+273 K) / 1.7atm

    = 2.25L

    Therefore,

    Work of expansion = - PV = - 1.7 atm x 2.25L

    = - 3.825atm L

    = - 3.825 x 101.325 J = - 387.57 J

    = - 390 J
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