Explanation:
N2 (g) + H2 (g) gives out NH3 (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.
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 NH3 a new stoichiometric coefficient.)
N2 (g) + 3H2 (g) gives out 2NH3 (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 mols.
So what do we do? Convert to
mols (remember the hint?).
28g N2 × 1 mol N2/ 2 × 14.007) g N2
= 0.9995 mol N2
At this point you don't even need to calculate the number of mols of H2 . Why? Because H2 is about 2 g/mol, which means we have over 10 mols of H2. We have 1 mol N2, and we need three times as many mols of H2 as we have
N2.
After doing the actual calculation you should realize that we have about 4 times as much H2 as we need. Therefore the limiting reagent is clearly N2.
Thus, we should yield 2×0.9995=1.9990 mols of NH3 (refer back to the reaction). So this is the second and last calculation we need to do:
1.9990 mol NH3 × 17.0307 g NH3/ 1 mol NH3
= 34.0444 g NH3
Hope it helpz~
