Question

Grams of `N_2` reacts with 84.29 grams of `H_2` in the chemical equation: `N_2 + H_2-> NH_3` How do you balance the chemical equation and determine the limiting reactant? How much `NH_3` is produced?

Answer 1

The stoichiometric equation is:

`1/2N_2(g) + 3/2H_2(g) rarr NH_3(g)`

Explanation:

Three equiv of hydrogen react with 1 equiv of nitrogen to give one equiv ammonia. Of course, I could double this reaction to remove the 1/2 coefficients, but the stoichiometry will remain the same.

If you have 84.29 g dihydrogen, there are `(84.29*cancelg)/(2.02*cancelg*mol^-1)` `=` `41.7` `mol` `H_2`. Stoichiometry requires 27.8 mol dinitrogen (`=` `?? g`). I think you have omitted some of the starting conditions in the problem