How many moles of ammonia can be made by reacting 5.70 moles of N2 with 4.95 moles of H2?
N2 (g) + 3 H2 (g) --> 2 NH3 (g
From the balanced equation:
N2 (g) + 3 H2 (g) ----------> 2 NH3 (g)
From this reaction, one mole of N2 react with three moles of H2. So, for the reaction of 5.7 moles of N2, number of mole require for H2 = 3 x 5.7 = 17.1 moles.
But here you have only 4.95 moles of H2 (reqired 17.1 moles)
Hence, the limiting reagent is H2 here.
Now, 3 mole of H2 provide 2 mole of NH3
So, after reaction of 4.95 moles of H2, produced NH3 = (2 x 4.95) / 3
= 9.9 / 3
= 3.3 moles of NH3
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