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?

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 )

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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?