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

Explanation

The balanced chemical equation for the reaction is :
`underset("1 mol")(N_(2)(g))+underset("3 mol")(3H_(2)(g))rarrunderset("2 mol")(2NH_(3)(g))`
Step I. To find the limiting reactant
Mass of `N_(2)` taken `= 50.0 kg = 50,000 g`
Moles of `N_(2)=("Mass of" N_(2))/("Molar mass of "N_(2))=(5000g)/((28"g mol"^(-1)))=1785.7mol`
Mass of `H_(2)` taken `= 10.0 kg = 10,000 g`
Moles of `H_(2)=(10,000g)/((2.0"g mol"^(-1)))=5000 mol`
From the above balanced equation, 1 mol of `N_(2)` reacts with 3 mol of `H_(2)`
`:. 1785.7` mol of `N_(2)` will react with` H_(2)=3xx1785.7 mol = 5355.9 mol` But we have only 5000 mol of `H_(2)`. Therefore, `H_(2)` is the limiting reactant.
Step II. To find the amount of `NH_(3)` formed.
3 moles of `H_(2)` from `NH_(3) =2` mol
5000 moles of `H_(2)` form `NH_(3)=5000 xx ((2 mol))/((3 mol))=3333.3 mol`
Mass of `NH_(3)` formed `= 3333.3 xx 17 = 56666.6 g = 56.67 kg`.