Question

How many grams of ammonia, `NH_3`, would be formed from the complete reaction of 4.50 moles hydrogen, `H_2` in the reaction `N_2 + 3H_2 -> 2NH_3`?

Answer 1

Approx. `54*g" ammonia"`

Explanation:

You have almost solved this problem yourself, because you have written a stoichiometrically balanced equation, which establishes equivalent mass:

`N_2(g) + 3H_2(g)rightleftharpoons 2NH_3(g)`

This tells us that `28*g` of dinitrogen combines with `6*g` dihydrogen to give `34*g` of ammonia. I take it that you know how I got these equivalent masses (if no, I am happy to explain).

You have `4.5*mol` dihydrogen, and thus, at most I can form `4.5*mol*H_2xx(2*mol" ammonia")/(3*mol" dihydrogen")` `=` `3*mol*" ammonia"`.

`"Mass of ammonia"` `=` `"moles "xx" molar mass"` `=` `3*molxx17.0*g*mol^-1` `=` `??g`