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

Explanation

A balanced equation for the above reaction is written as follows :
Calculation of moles :
`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)`
moles of `N_(2)=50.0kgN_(2)xx(1000gN_(2))/(1kgN_(2))xx(1molN_(2))/(28.0gN_(2))=17.86xx10^(2)mol`
moles of `H_(2)=10.00kgH_(2)xx(1000gH_(2))/(1kgH_(2))xx(1molH_(2))/(2.016gH_(2))=4.96xx10_(3)mol`
According to the above equation, 1 mol `N_(2)(g)` requires 3 mol `H_(2)(g)`, for the reaction. Hence, for `17.86xx10^(2)` mol of `N_(2)`, the moles of `H_(2)(g)` required would be
`17.86xx10^(2)molN_(2)xx(3molH_(2)(g))/(1molN_(2)(g))=5.36xx10^(3)molH_(2)`
But we have only `4.96xx10^(3)` mol `H_(2)`. Hence, dihydrogen is the limiting reagent in this case. So `NH_(2)(g)` would be formed only from that amount of available dihydrogen i.e., `4.96xx10^(3)mol`
Since 3 mol `H_(2)(g)` gives 2 mol `NH_(3)` (g)
`4.96xx10^(3)molH_(2)(g)xx(2molNH_(3)(g))/(3molH_(2)(g))=3.30xx10^(3)molNH_(3)(g)`
`3.30xx10^(3)molNH_(3)`(g) is obtained.
If they are to be converted to grams, it is done as follows :
1 mol `NH_(3)`(g) = 17.0 g `NH_(3)`(g)
`3.30xx10^(3)molNH_(3)(g)xx(17.0gNH_(3)(g))/(1molNH_(3)(g))`
`=3.30xx10^(3)xx17gNH_(3)(g)`
`=56.1xx10^(3)gNH_(3)`
`=56.1kgNH_(3)`