Dinitrogen and dihydrogen react with each other to produce ammonia according to the following equation : `N_2(g) + 3H_2(g) to 2NH_3(g)`
(i) Calculate the mass of ammonia produced if `2.00 xx 10^3` g dinitrogen react with `1.00 xx 10^(3)` g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?

Explanation

`1 mol` of `N_(2)`, i.e., `28 g` reacts with `3` mole of `H_(2)`, i.e., `6 g` of `H_(2)`.
`:. 2000 g` of `N_(2)` will react with `H_(2)=6/28xx200 g =428.6 g`. Thus, `N_(2)` is the limiting reagent while `H_(2)` is the excess reagent.
`2 mol` of `N_(2)`, i.e., `28 g` of `N_(2)` produces `NH_(3)= 2 mol`
`=34 g`
Therefore, `2000 g` will produces `NH_(3)=34/28xx2000 g`
`=2428.57 g`.
b. `H_(2)` will remain unreacted.
c. Mass left unreacted `=1000 g-428.6 g=571.4 g`