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31 January, 23:57

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

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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Answers (1)
  1. 1 February, 00:14
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    (a) Theoretical Yield = 10.50 g

    (b) %age yield = 27.33 %

    Explanation:

    (a)

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

    N₂ + 3 H₂ → 2 NH₃

    Step 1: Calculate moles of N₂ as;

    Moles = Mass / M/Mass

    Moles = 9.75 g / 28.01 g/mol

    Moles = 0.348 moles

    Step 2: Calculate moles of H₂ as;

    Moles = Mass / M/Mass

    Moles = 1.86 g / 2.01 g/mol

    Moles = 0.925 moles

    Step 3: Find 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 Hydrogen is limiting reagent. Therefore, H₂ will control the final yield.

    Step 4: Calculate 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: Calculate theoretical yield of Ammonia as,

    Theoretical Yield = Moles * M. Mass

    Theoretical Yield = 0.616 mol * 17.03 g/mol

    Theoretical Yield = 10.50 g

    (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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