Question

nitrogen and hydrogen combine at high temperature, in the presence of a catalyst, to produce ammonia. N2(g)+3H2(g) > 2NH3(g). assume 0.270 mol of N2 and 0.877 mol of H2 are present initially. Questions- 1. after complete reaction, how many moles of ammonia are produced? 2. how many moles of H2 remain? 3. how many moles of N2 remain? 4. what is the limiting reactant(hydrogen or nitrogen)?

Answer 1

The reaction is as follows:

N₂(g) + 3H₂(g) <----> 2NH₃(g)

According to reaction, 1 mol of N₂(g)------> 3 mol of H₂(g)

Therefore, the stiochiometric ratio is moles of N₂(g)/moles of H₂(g) =1/3 = 0.33

But, the actual ratio of moles of N₂(g)/moles of H₂(g) = 0.27/0.877 = 0.3033

N₂ is limiting reactant, it reacts completely and 0 moles of N₂ remains unreacted.

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According to the limiting reactant,moles of NH₃ produced

1 mol of N₂(g) ------> 2 mol of NH₃(g)

0.27 mol of N₂(g) ------> mol of NH₃(g)

moles of NH₃ produced =2*0.27=0.54 mol

Therefore, moles of  NH₃ produced =0.54 mol

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According to the limiting reactant,moles of NH₃ produced

1 mol of N₂(g) ------> 3 mol of H₂(g)

0.27 mol of N₂(g) ------>

moles of H₂ reacted =3*0.27=0.81 mol

Excess of H₂ =initial H₂ fed- H₂ reacted = 0.877-0.81 = 0.067 mol

Therefore, moles of   H₂ remain= 0.067 mol

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Moles of N₂ remained is 0 as it is reacted completely.