Question

How many grams of `NH_3` can be produced from the reaction of 28 g of `N_2` and 25 g of `H_2`?

Answer 1

Well, what is the reaction itself? You can always start there.

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

Now balance it. You have two reactants with compositions involving a single element, which makes it very easy to keep track of how much is on each side. I would balance the nitrogens, and then the hydrogens.

(If you balance the hydrogen reactant with a whole number first, I can guarantee you that you will have to give `"NH"_3` a new stoichiometric coefficient.)

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

The stoichiometric coefficients tell you that if we can somehow treat every component in the reaction as the same (like on a per-mol basis, hinthint), then one "[molar] equivalent" of nitrogen yields two [molar] equivalents of ammonia.

Luckily, one `"mol"` of anything is equal in quantity to one `"mol"` of anything else because the comparison is made in the units of `"mol"`s. So what do we do? Convert to `"mol"`s (remember the hint?).

`"28" cancel("g N"_2) xx ("1 mol N"_2)/((2xx14.007)cancel("g N"_2)) = "0.9995 mol N"_2`
(how convenient... pretty much `"1 mol"`)

At this point you don't even need to calculate the number of `\mathbf("mol")`s of `\mathbf("H"_2)`. Why? Because `"H"_2` is about `"2 g/mol"`, which means we have over `"10 mol"`s of `"H"_2`. We have `"1 mol N"_2`, and we need three times as many `"mol"`s of `"H"_2` as we have `"N"_2`.

After doing the actual calculation you should realize that we have about `4` times as much `"H"_2` as we need. Therefore the limiting reagent is clearly `"N"_2`.

Thus, we should yield `2xx0.9995 = "1.9990 mol"`s of `"NH"_3` (refer back to the reaction). So this is the second and last calculation we need to do:

`"1.9990" cancel("mol NH"_3) xx ("17.0307 g NH"_3)/(cancel("1 mol NH"_3)) = color(blue)("34.0444 g NH"_3)`