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28 May, 20:09

If 15.0g of nitrogen reacts with 15.0g of hydrogen, 10.5g of ammonia os produced. What is the percent yield of this reaction.

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  1. 28 May, 20:25
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    Percent yield = 57.2%

    Explanation:

    Given data;

    Mass of nitrogen = 15.0 g

    Mass of hydrogen = 15.0 g

    Mass of ammonia produced = 10.5 g

    Percent yield = ?

    Solution:

    Chemical equation:

    N₂ + 3H₂ → 2NH₃

    Number of moles of nitrogen:

    Number of moles = mass / molar mass

    Number of moles = 15 g / 28 g/mol

    Number of moles = 0.54 mol

    Number of moles of hydrogen:

    Number of moles = mass / molar mass

    Number of moles = 15 g / 2 g/mol

    Number of moles = 7.5 mol

    Now we compare the moles ammonia with hydrogen and nitrogen

    N₂ : NH₃

    1 : 2

    0.54 : 2/1 * 0.54 = 1.08 mol

    H₂ : NH₃

    3 : 2

    7.5 : 2/3*7.5 = 5 mol

    The number of moles of ammonia produced by nitrogen are less so it will limiting reactant.

    Theoretical yield:

    Mass = number of moles * molar mass

    Mass = 1.08 mol * 17 g/mol

    Mass = 18.36 g

    Percent yield:

    Percent yield = actual yield / theoretical yield * 100

    Percent yield = 10.5 g / 18.36 g * 100

    Percent yield = 57.2%
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