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Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia. N 2 ( g ) + 3 H 2 ( g ) ⟶ 2 NH 3 ( g ) Assume 0.180 mol N2 and 0.575 mol H2 are present initially. After complete reaction, how many moles of ammonia are produced

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joseaaronlara

Answer:

The answer to your question is 0.36 moles of NH₃

Explanation:

Data

moles of N₂ = 0.180

moles of H₂ = 0.575

moles of NH₃ = ?

- Balanced chemical reaction

             N₂ (g)  +  3H₂ (g)  ⇒  2NH₃ (g)

Process

1.- Calculate the limiting reactant

Theoretical proportion    N₂/H₂ = 1/3 = 0.333

Experimental proportion N₂/H₂ = 0.18 / 0.575 = 0.31

Conclusion

As the experimental proportion was lower than the theoretical proportion, we conclude that the N₂ is the limiting reactant.

2.- Calculate the moles of Ammonia

                    1 mol of N₂  ----------------- 2 moles of NH₃

                    0.180 moles ----------------  x

                    x = (0.180 x 2) / 1

                    x = 0.36 moles of NH₃

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