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) = `50.0" kg" N_(2)xx(1000 g N_(2))/(1kg N_(2))xx(1 "mol"N_(2))/(28.0 g N_(2))`
17.86 xx `10^(2)` mol
Moles of `H_(2)`
10.0 kg `H_(2)` xx `(1000 g H_(2))/(1kg H_(2))xx(1"mol"H_(2))/(2.016 g H_(2))`
4.96 xx `10^(3)` mol Accerding to the above equation 1 mol `N_(2)(g)` requires 3 mol `H_(2)(g)` for the reaction . Hence for 17.86 xx `10^(2)` mol `N_(2)` the moles of `H_(2)(g)` required would be 17.86 xx `10^(2)` mol `N_(2)` xx `(3"mol"H_(2)(g))/(1"mol"N_(2)(g))`
=5.36 xx `10^(3)` mol `H_(2)` But we have only 4.96 xx `10^(3)` mol `H_(2)` Hence dihydrogen is the limiting reagent in this case. So `NH_(3)(g)` Would be formed only from that amount of avilable dihydrogen i.e.,4.96 xx `10^(3)`mol. Since 3 mol `H_(2)(g)` gives 2 mol `NH_(3)(g)`
4.96 xx `10^(3)` mol `H_(2)(g)x(2"mol"NH_(3)(g))/(3"mol"H_(2)(g))`
=3.30 xx `10^(3)" mol" NH_(3)(g) 3.30xx10^(3)" mol"NH_(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)"mol"NH_(3)(g)xx(17.0 g NH_(3)(g))/(1"mol"NH_(3)(g))`
=`3.30xx10^(3)xx17 g NH_(3)(g)`
= `56.1xx10^(3)gNH_(3)(g)`
`56.1 kg NH_(3)`