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16 May, 04:59

1.86 g H2 is allowed to react with 9.75 g N2, producing 2.87g NH3.

A. What is the theoretical yield in grams for this reaction under the given conditions?

B. What is the percent yield for this reaction under the given conditions?

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  1. 16 May, 05:57
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    (a) Theoretical Yield = 10.50 g

    (b) %age yield = 27.33 %

    Explanation:

    Answer-Part - (a)

    The balance chemical equation for the synthesis of Ammonia is as follow;

    N₂ + 3 H₂ → 2 NH₃

    Step 1: Calculating moles of N₂ as;

    Moles = Mass / M/Mass

    Moles = 9.75 g / 28.01 g/mol

    Moles = 0.348 moles of N₂

    Step 2: Calculating moles of H₂ as;

    Moles = Mass / M/Mass

    Moles = 1.86 g / 2.01 g/mol

    Moles = 0.925 moles

    Step 3: Finding Limiting reagent as;

    According to equation,

    1 mole of N₂ reacts with = 3 moles of H₂

    So,

    0.348 moles of N₂ will react with = X moles of H₂

    Solving for X,

    X = 3 mol * 0.348 mol / 1 mol

    X = 1.044 mol of H₂

    It shows that to consume 0.348 moles of N₂ completely we require 1.044 mol of Hydrogen while, as given in statement we are only provided with 0.925 moles of H₂ hence, hydrogen is limiting reagent. Therefore, H₂ will control the final yield.

    Step 4: Calculating moles of Ammonia as,

    According to equation,

    3 mole of H₂ produces = 2 moles of NH₃

    So,

    0.925 moles of H₂ will produce = X moles of NH₃

    Solving for X,

    X = 2 mol * 0.925 mol / 3 mol

    X = 0.616 mol of NH₃

    Step 5: Calculating theoretical yield of Ammonia as,

    Theoretical Yield = Moles * M. Mass

    Theoretical Yield = 0.616 mol * 17.03 g/mol

    Theoretical Yield = 10.50 g

    Answer-Part - (b)

    %age yield = Actual Yield / Theoretical Yield * 100

    %age yield = 2.87 g / 10.50 g * 100

    %age yield = 27.33 %
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