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 situtation.

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

1.	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 situtation.
Answer» Solution :The BALANCED chemical equaiton for the reaction is `N_(2)(g)+3H_(2)(g)rarr 2NH_(3)(g)` STEP 1. To convert the GIVEN amounts into moles Mass of `N_(2)` taken `=50.0kg = 50,000g` `"Moles of "N_(2)=("Mass of N"_(2))/("Molar mass of N"_(2))=(5000g)/("28 g mol"^(-1))="1785.7 moles"` Mass of `H_(2)` taken = 10.0 kg= 10,000 g Step 2. To identify the limiting reagent From the above balanced equation, 1 mole of `N_(2)` reacts with 3 moles of `H_(2)` `therefore"1785.7 moles of N"_(2)" will react with "H_(2)=3xx1785.7" mol = 5355.9 moles"` But we have only 4960.3 moles of `H_(2)`. Hence, `N_(2)` is not the limiting reactant. Testing in the reverse manner, 3 moles of `H_(2)` react with 1 moles of `N_(2)` To calculate the amount of `NH_(3)` formed As the amount of product formed depends UPON the limiting reagent, hence, we calculate `NH_(3)` formed as follows: `"3 moles of "H_(2)" from "NH_(3)="2 moles"` `therefore"4960.3 moles of "H_(2)" will form "NH_(3)=(2)/(3)xx4960.3mol=" 3306.9 moles"` Molar mass of `NH_(3)=17gmol^(-1)` `therefore"Mass of "NH_(3)" formed "=3306.9xx17g=56217g=56.217kg`