Question

Ammonia is produced in the reaction `3H_2 + N_2 -> 2NH_3`. What mass of `NH_3` could be produced if 12.5 `H_2` reacts with excess nitrogen?

Answer 1

We assess the reaction...`1/2N_2(g)+3/2H_2(g) rarr NH_3(g)`

Explanation:

We ASSUME `(i)` `12.5*g` dihydrogen gas, and `(ii)` QUANTITATIVE reaction with stoichiometric dinitrogen. (And the second assumption is a bit unwarranted).

`"Moles of dihydrogen"=(12.5*g)/(2.016*g*mol^-1)=6.20*mol`...and thus `2.07*mol` of dinitrogen are required. We get `4.14*mol` ammonia gas, a mass of `4.14*molxx17.01*g*mol^-1=70.31*g`...