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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
- FernLv 75 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%