Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
`N_(2)(g)+3H_(2)(g)rarr 2NH_(3)(g)`
a. Calculate the mass of ammonia produced if `2.00xx10^(3) g` dinitrogen reacts with `1.00xx10^(3) g` of dihydrogen.
b. Will any of the two reactants remain unreacted?
c. If yes, which one and what would be its mass?

Explanation

The balanced chemical equation is:
`N_(2)(g) + 3H_(2)(g) Moles of `N_(2) =(2.00 xx 10^(3))/28=71.43` mol
Moles of `H_(2) = (1.00 xx 10^(3))/2 = 500`mol
According to above equation, 1 mol of `N_(2)` require 3 mol of `H_(2)`.
71.43 mol of `N_(2)` will require `H_(2) =3 xx 71.43`
=214.29 mol
But moles of `H_(2)` actually present = 500 mol
`H_2` is in excess and will remain unreacted and `N_(2)` is limiting reagent.
(i) 1 mol of `N_(2)` react with `H_(2)` to form `NH_(3)`
`=2/1 xx 71.43`
=142.86 mol
Mass of`NH_(3)` produced = `142.86 xx 17 = 2428.6` g
(ii) Hydrogen will remain unreacted.
(iii) Moles of `H_2` initially taken = 500 mol
Moles of `H_2` reacting = 214.29 mol
Moles of `H_2` remaining unreacted
`= 500 - 214.29 = 285.71` mol
Mass of `H_(2)` left unreacted = `285.71 xx 2 = 571.42` g