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

Moles of `N_(2) =("Mass in g")/("molar mass")`
`=(5.0 xx 10^(3))/28 = 1.78 xx 10^(3)` mol
Moles of `H_(2) =("Moles in g")/("Molar mass")`
`=(10 xx 10^(3))/2 = 5.0 xx 10^(3)` mol
`underset(1 "mol")(N_(2)(g)) + underset(3"mol")(3H_(2)(g)) According to the equation 1 mol of `N_(2)`(g) requires 3 mol of `H_(2)(g)` for the reaction:
`1.768 xx 10^(3)` mol of `N_(2)(g)` will require `H_(2)(g)`
`=3 xx 1.786 xx 10^(3)`
`=5.36 xx 10^(3)` mol
Therefore, `N_(2)` is in excess and dihydrogen is limiting reagent.
Now, 3 mol of `H_(2)`(g) gives `NH_(3)(g)` =2 mol
`5.0 xx 10^(3)` mol of `H_(2)`(g) will give `NH_(3)`(g)
`=2/3 xx 5.0 xx 10^(3)`
`=3.3 xx 10^(3)` mol
Mass of `NH_(3)` (g) formed = Moles x Molar mass
`=3.3 xx 10^(3) xx 17`
=56.1 kg