Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation :
`N_(2(g))+H_(2(g)) rarr 2NH_(3(g))`
(i) Calculate the mass of ammonia produced if `2.00xx10^(3)g` dinitrogen reacts with `1.00xx10^(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

`{:(N_(2(g)),+,3H_(2),rarr,2NH_(3(g)),),(1" mol",,3 " mol",,2 " mol",),(28 gm,,6 gm,,34 gm,):}`
28 gm `N_(2)` react with 6 gm `H_(2)`
`:. 2000 gm N_(2) rarr (?)`
`= (2000 xx 6)/(28) = 428.57 g H_(2)`
(i) Here, `N_(2)` is limiting reagent So, created volume of ammonia.
From 28 gm `N_(2)`, 34 gm `NH_(3)` create.
`:.2000 gm N_(2) rarr (?)`
`= (2000xx34)/(28) = 2428.57 g NH_(3)`
(ii) Here, `N_(2)` is limiting reagent So, `H_(2)` become excess. (unreacted)
(iii) Remaining mass of `H_(2)` gas `= 1000-428.57`
`= 571.43g`