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
#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.
#"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
At this point you don't even need to calculate the number of
After doing the actual calculation you should realize that we have about
Thus, we should yield
#"1.9990" cancel("mol NH"_3) xx ("17.0307 g NH"_3)/(cancel("1 mol NH"_3)) = color(blue)("34.0444 g NH"_3)#