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5 August, 02:19

As a result of the reaction, 34.1 grams of ammonia (NH3) was produced. What volume of hydrogen gas at 20.0°C and 1.5 atm was required for this mass of ammonia to be produced, assuming sufficient nitrogen to react?

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  1. 5 August, 05:35
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    The volume of H₂ produced is 32.03L

    Explanation:

    The reaction is this:

    N₂ + 3H₂ → 2NH₃

    Ratio between H₂ and ammonia is NH₃ is 2:3 so the rule of three will be

    2 moles from ammonia are produced by 3 moles of hydrogen

    My moles of ammonia were produced by ...

    I have the mass of the produced ammonia so:

    Mass / Molar mass = Moles

    34.1 g / 17 g/m = 2 moles

    (2.3) / 2 = 3 moles of Hydrogen were necessary to produce 34.1 g

    Apply the Ideal Gases Law equation:

    P. V = n. R. T

    1.5 atm. V = 2 moles. 0.082. 293K)

    V = (2 moles. 0.082. 293K) / 1.5atm → 32.03L
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