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 react with 3 mol of `H_(2)`, i.e., 6 of `H_(2)`. 2000 g of `N_(2)` will react with `H_(2) = (6)/(28) xx 2000g = 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)` produce `NH_(3) = 2` mol = 34 g 2000 g of `N_(2)` will produce `NH_(3) = (34)/(28) xx 2000g = 2428.57 g`
(B) `H_(2)` will remain unreacted
(C ) Mass left unreacted = 1000 g – 428 .6 g = 571.4 g