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16 April, 02:31

The decomposition of 128.4g ammonium nitrate yields how many liters of water at 1.89atm and 314K?

NH₄NO₃ (s) →N₂O (g) + 2H₂O (g)

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  1. 16 April, 03:29
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    First, we calculate the moles of ammonium nitrate:

    Moles = mass / molecular mass

    Moles = 128.4 / 80

    Moles = 1.60

    From the equation, it is visible that the molar ratio between ammonium nitrate and water is 1 : 2. The moles of water formed will be:

    1.6 * 2 = 3.2 moles

    Next, we may apply the ideal gas law equation to find the volume:

    PV = nRT

    V = nRT/P

    V = (3.2 * 0.082 * 314) / 1.89

    V = 43.6 liters

    43.6 liters of water will be produced
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