50.0 kg of N_(2)(g) and 10.0 kg of H_(2)(g) are mixed to produce NH_(2)(g). Calculate the NH_(2)(g) formed. Identify the limiting reagent in the production of NH_(3) in this situation.

50.0 kg of N_(2)(g) and 10.0 kg of H_(2)(g) are mixed to produce NH_(2

1.	50.0 kg of N_(2)(g) and 10.0 kg of H_(2)(g) are mixed to produce NH_(2)(g). Calculate the NH_(2)(g) formed. Identify the limiting reagent in the production of NH_(3) in this situation.
Answer» Solution :The gasesous phase chemical equation for the production of ammonia is GIVEN as `N_(2)+3H_(2) to 2NH_(3)` 1 mole of `N_(2)="3 moles of "H_(2)` 28kg of `N_(2)=(2 XX 3)" kg of "H_(2)` 46.67 kg of `N_(2)` =10 kg of `H_(2)` 50kg of `N_(2)=10.71"kg of "H_(2)` Hence, HYDROGEN is the limiting reagent. 3 moles of `H_(2)="2 moles of "NH_(3)` 10 kg of `H_(2)=?` Maximum amount of ammonia produced is `(2 xx 17 xx 10)/(3 xx 2)=56.6kg`