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.

Explanation

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`