6.16 x 10(2) kg of nitrogen are reacted with 1.75 x 10(2) kg hydrogen under conditions of high temperature in the presence of a catalyst to produce 1.12 x 10(2) kg ammonia [N(2) +H(2) -> NH(3)] Percent yield?


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John McCann Profile
John McCann answered

I will do this top one to show you how it goes. You do the other problem then. Balanced equation.

2N2 + 3H2 --> 2NH3

I will change the two reactants mass into grams as the periodic table has g/mol masses. I will keep the actual yield in kg and switch back when through for the last step.

6.16 X 10^2 kg N2 = 616,000 grams N2

1.75 X 10^2 kg H2 = 175,000 grams H2

Find limiting reactant by finding moles of reactant.

616,000 grams N2 (1 mol N2/28.02 grams) = 21,984.297 mols N2

175,000 grams H2 (1 mol H2/2.016 grams) = 86,805.556 mols H2

I suspect N2 of limiting.

21,984.297 moles N2 (3 mol H2/2 mol N2) = 32,976446 mols

You do not have that many mols N2, so N2 drives the reaction. ( do it the other way and see that you have more than enough mols H2 )


21,984.297 mols N2 (2 mol NH3/2 mol N2)(17.034 grams/1 mol NH3)

= 374,481 grams NH3 = 3.74 X 10^2 kg ammonia theoretical production.

1.12 X 10^2 kg actual production/3.74 X 10^2 theoretical production

= 0.299465 * 100

= 30% yield

===================Final answer.

Sounds reasonable for the Born-Haber process. Do you know why the process is at high temperature?

2 People thanked the writer.
Tom  Jackson
Tom Jackson commented
Is this your only post under "anonymous?"
John McCann
John McCann commented
Most questions come in as anonymous in this section

What is your point?

By the way, this is science, not delusion. I doubt that you can understand anything by the articles ( the, if, and ) above.

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