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

A balanced equation for the above reaction is written as follows,
`N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g))`.
Mole of `N_(2)=50.0 kg N_(2) xx (1000 g N_(2))/(1 kg N_(2)) xx (1 "mol" N_(2))/(28.09g N_(2))`
`= 17.86 xx 10^(2)` mol
Mole of `H_(2) = 10.00 kg H_(2) xx (1000 g H_(2))/(1 kg H_(2)) xx (1 "mol" H_(2))/(2.016 g H_(2))`
`= 4.96xx10^(3)` mol
According to the equation , 1 mol `N_(2(g))` required 3 mol `(H_(2))`. For the reaction.
Hence, for `17.86xx10^(2)` mol of `N_(2(g))` , the moles of `H_(2(g))` required would be :
`= 17.86 xx 10^(2)` mol `xx(3 "mol"H_(2(g)))/(1 "mol"N_(2(g)))`
`= 5.36xx10^(3)` mol `H_(2(g))`
But we have only `4.96xx10^(3)` mol `H_(2)`. Hence diphydrogen is the limiting reagent in this case. So, `NH_(3(g))` would be formed only from that amount of available dihydrogen , i.e `4.96xx10^(3)` mol `H_(2(g))`.
Since 3 mol `H_(2(g)), 2 mol NH_(3(g))`
`4.96xx10^(3) mol H_(2(g)) xx (2 mol NH_(3(g)))/(3 mol H_(2(g)))`
`= 3.30xx10^(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 mol NH_(3(g)))/(1 mol NH_(3(g)))`
`= 3.30xx10^(3) xx17 g NH_(3(g))`
`= 56.1 xx 10^(3) g NH_(3)`
`= 56.1 kg NH_(3(g))`