CHEMISTRY 123

Question

1.38 g H2 is allowed to react with 10.2 g N2, producing 2.35 g NH3.
Part A
What is the theoretical yield in grams for this reaction under the given conditions?
Express your answer to three significant figures and include the appropriate units.
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Answer 1

Write the balanced equation

--  3H₂(g) + N₂(g) ===> 2NH₃(g)

 

Determine the limiting reactant

--  moles H₂ = 1.38 g x 1 mole/2 g = 0.69 moles

--  moles N₂ = 10.2 g x 1 mole/28 g = 0.36 moles

--  H₂ is limiting as determined from mole ratio of 3:2 in the balanced equation

 

Determine moles NH3 theoretically formed

-  0.69 moles H₂ x 2 moles NH₃/3 moles H₂ = 0.46 moles

 

Determine mass of NH₃ theoretically formed

--  0.46 moles NH₃ x 17 g/mole = 7.82 g (three significant figures and units are grams)