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

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`