How many moles of #NH_3# can be produced from 21.0 mol of #H_2# and excess #N_2#?

Answer 1

Clearly, #14*mol# of ammonia could be produced.

The equation for stoichiometry:

#N_2(g) + 3H_2(g)rarr2NH_3(g)#

According to the stoichiometric equation, every mol of nitrogen requires three mol of dihydrogen.

Given that we have #21.0*mol" dihydrogen"#, #14*mol# of ammonia could be produced given stoichiometric reaction. Of course, in this reaction, stoichiometric yields are unthinkable, but industrial procedures can recycle the unreacted dinitrogen and dihydrogen and achieve acceptable turnovers. That ammonia is condensable, whereas the reactant gases are much less so, allows recycling.
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Answer 2

10.5 moles of NH₃ can be produced from 21.0 mol of H₂ in excess N₂.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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