Question

. Ammonia, NH3, is a typical ingredient in household cleaners. It is produced through a combination reaction involving N2(g) and H2(g). If 24.0 mol of H2(g) react with excess N2(g), how many moles of ammonia are produced? Show your work.

Answer 1

Let us first write the balanced reaction of formation of \(NH_3\) :

\(N_2\) + 3 \(H_2\) \(\rightarrow\) 2 \(N H_3\)

(all are in gaseous form)

Answer 2

It is given to you that :

24 mol. of \(H_2\) (g) react with excess \(N_2\) (g)

So, using stoichiometry , we get :

\(24 mol. H_2 \times \cfrac{2 mol. NH_3}{3 mol. H_2} = \) x mol. \(NH_3\)

x = no. of moles of \(NH_3\) produced

Solve it for x.

Answer 3

Refer to this link too : http://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm

Answer 4

im completely lost on this whole topic in general

Answer 5

@mathslover pretty much did the entire problem for you. you just have to solve

\[24 \text{ mol. H}_2 * \frac{2 \text{ mol. NH}_3}{3\text{ mol. H}_2} = x \text{ mol. NH}_3\]That looks forbidding at first, but most of it is just the units, and they cancel:

\[24 \cancel{\text{ mol. H}_2 }* \frac{2 \text{ mol. NH}_3}{3\cancel{\text{ mol. H}_2}} = x \text{ mol. NH}_3\]
\[24*\frac{2}{3}\text{ mol. NH}_3 = x \text{ mol. NH}_3\]
\[24*\frac{2}{3}\cancel{\text{ mol. NH}_3 }= x \cancel{\text{ mol. NH}_3}\]\[24*\frac{2}{3} = x\]Surely you can do that!

Answer 6

16

Answer 7

Yes, so there will be 16 moles of NH3 produced. Let's check it out:

24 moles of H2 = 48 moles of H atoms
unlimited N2 is provided
each mole of NH3 requires 3 moles of H atoms and 1 mole of N atoms
we can make 48/3 = 16 moles of NH3 from our 48 moles of H atoms and unlimited N atoms