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?

Answer

(i) `2.43 xx10^(3) g " " (ii) "yes "`
(iii) Hydrogen will remain unreacted ; `5.72 xx10^(2)` g

Explanation

(i) Balancing the given chemical equation,
`N_(2(g)+3H_(2(g))to 2 NH_(3(g))`
From the equation , 1 mole (28 g )of dinitrogen reacts with 3 mole (6 g) of dihydrogen to Give 2 mole (34 g) of ammonia .
`implies 2.00 xx10^(3)` g of dinitrogen will react with `(6 g)/(28 g)xx2.00 xx10^(3) g` dihydrogen i.e .,
`2.00xx10^(3)` g dinitrogen will dinitrogen will react with 428.6 g of dihydrogen .
Given ,
amount of dihydrogen `=1.00 xx10^(3)`g
Hence ,`N_(2)` is the limiting reagent .
`therefore` 28 g of `N_(2)` prouces 35 g of `NH_(3)`
Hence , mass of ammoina produced by 2000 g of `N_(2) =(34 g)/(28 g)xx2000g`
=2428.57 g
(ii) `N_(2)` is the limiting reagent and `H_(2)` is the excess reagent .Hence ,`H_(2)` will remain unreacted.
(ii) Mass of dihydrogen left unreacted `=1.00 xx10^(3) g-428.6 g`
=571.4 g