Question

In the equation `N_2(g) + 3H_2(g) -> 2NH_3(g)`, 8.5 grams of hydrogen and 15.8 grams of nitrogen react to form ammonia. How many grams of ammonia are produced?

Answer 1

Find the limiting reagent.

Explanation:

Moles of `N_2` reactant `=` `(15.8*g)/(28.02*g*mol^-1)` `=` `0.564` `mol`.

Moles of `H_2` reactant `=` `(8.5*g)/(2.02*g*mol^-1)` `=` `4.21` `mol`.

Dihydrogen is the reagent in excess (clearly!). Given that dinitrogen is the limiting reagent, at most I can form `1.13` mol ammonia gas (i.e. 2 equiv with respect to dinitrogen).

If I have formed `1.13` `mol` ammonia, this is `1.13` `cancel("mol")xx17.0*g*cancel("mol"^-1)` `=` `??g`

You might already know that this is probably the most important reaction on the planet. An industrialist would have an orgasm if he could achieve such yields.