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

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

the correct answer is:

a. formation of precipitation

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