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

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)#