Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation.
`N_(2)(g)+H_(2)(g)to2NH_(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

i) The balanced equation for the reaction between dihydrogen and dinitrogen is `underset(28)(N_(2))+underset(6)(3H_(2))hArrunderset(34)(2NH_(3))`
When 28 g of `N_(2)` react with 6 g of `H_(2)` produce 34 g of `NH_(3)`
Moles of `N_(2)=(2xx10^(3))/28=71.4` moles
Moles of `H_(2)=(1xx10^(3))/2=5xx10^(2)` moles or 500 mole.
1 mole of `N_(2)` can react with 3 moles of `H_(2)`
71.4 moles of `N_(2)` can react = `(71.4xx3)/1` = 214.2 moles of `H_(2)`
Here 1 mole of `N_(2)` can produce 34 g of `NH_(3)`
71.4 mole of `N_(2)` can produce
`71.4xx34=2427.6` gm.
iii) Here No. of moles of `H_(2)` are more than required
The no. of moles of `H_(2)` unreacted
= 500 - 214.2 = 285.8
The amount of hydrogen left = `285.8xx2` = 571.6 gm.