Question

`50.0kg` of `N_(2)(g)` and `10.0 kg` of `H_(2)(g)`. Are mixed to produce `NH_(3)(g)`. The limiting reagent and amount of `NH_(3)` formed are

Answer 1

`{:(N_(2)(g),+3H_(2)(g),hArr2NH_(3)(g)),((2xx14)g,6g,2(14+3)g),(or,or,or),(28kg,6kg,34kg):}`
According to the balanced chemical reaction, 28kg of `N_(2)` reacts with `6kg` of `H_(2)` to give 34 kg `Nh_(3)`. Thus, 6kg of `H_(2)` required `-= 28 kg` of `N_(2)`
10 kg of `H_(2)` requires `-= (28 xx 10)/(6) = 46.6 kg` of `N_(2)`
Hence, `H_(2)` is the limiting reagent, since it is completely consumed in the reaction and `H_(2)` will decide the amount of product formed.
Thus, `6kg of H_(2) =-= 34 kg` of `NH_(3)`
`10 kg` of `H_(2) -= (34 xx 10)/(6) = 56.6 kg` of `NH_(3)`