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1.07 g H2 is allowed to react with 9.81 g N2, producing 2.49 g NH3.?

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

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

I need to understand how we arrive at this answer not just the answer, please and thank you.

1 Answer

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  • Fern
    Lv 7
    5 years ago

    N2 + 3H2 ==⇒ 2NH3

    1.07 g H2 x 1mole H2/2.0 g x 2 moles NH3/3moles H2 = 0.357 moles NH3

    9.81 g N2 x 1mole N2/28.0 g x 2 moles NH3/1 mole N2 = 0.700 moles NH3

    One cannot obtain more than 0.357 moles of NH3. Thus, H2 is

    the limiting reactant.

    Theoretical yield of NH3:

    0.357 moles NH3 x 17.0 g NH3/1mole NH3= 6.07 grams

    Percent Yield = amount obtained/theoretical * 100

    Percent yield = 2.49 g NH3/6.07 g NH3 x 100 = 41.0%

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