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26 August, 11:25

Calculate the total volume of gas (at 119? c and 731 mmhg produced by the complete decomposition of 1.71 kg of ammonium nitrate.

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  1. 26 August, 11:32
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    The reaction is:

    NH4 (NO3) (s) ⇄ N2O (g) + 2 H2O (g)

    This means that 1 mol of NH4 (NO3) s produces 3 moles of gases.

    Now find the number of moles in 1.71 kg of NH4 (NO3)

    Molar mass = 2*14g/mol + 4 * 1g/mol + 3*16g/mol = 80 g/mol

    # moles = mass / molar mass = 1710 g / 80 g/mol = 21.375 mol of NH4 (NO3)

    We already said that every mol of NH4 (NO3) produces 3 moles of gases, then the number of moles of gases produced is 3 * 21.375 = 64.125 mol

    Now use the equation for ideal gases to fin the volume

    pV = nRT = > V = nRT / p = (64.125 mol) (0.082atm*liter / K*mol) * (119 + 273) K / (731mmHg * 1 atm/760mmHg) =

    V = 2143.01 liters
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