50.0 kg of `N_(2) (g)` and 10.0 kg of `H_(2)` (g) are mixed, to produce `NH_(3)(g)`. Calculate the `NH_(3)(g)`. Formed. Identify the limiting reagent in the production of `NH_(3)` in this situation.

Explanation

`N_(2)(g) + 3H_(2)(g) No. of moles of `N_(2) = 50.0 kg xx (1000 g)/(1 kg) xx (1 mol)/(28g)`
`=17.86 xx 10^(2)` mol.
No. of moles of `H_(2) = 10.00 kg xx (1000 kg)/(1 kg) xx (1 mol)/(2.016 g)`
`= 4.96 xx 10^(3)` mol.
According to the equation.
`17.86 xx 10^(2)` mol of `N_(2)` will react with
`=3/1 xx 17.86 xx 10^(2) = 5.36 xx 10^(3)` mol
Since number of moles of `H_(2)` are `4.96 xx 10^(3)`
`therefore` dihydrogen is the limiting reagent.