50.0 kg of N_(2)(g) and 10.0 kgof 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.

50.0 kg of N_(2)(g) and 10.0 kgof H_(2)(g) are mixed to produce NH_(3)

1.	50.0 kg of N_(2)(g) and 10.0 kgof 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.
Answer» Solution :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`.