The illustration to the left represents a mixture of nitrogen ( blue ) and hydrogen ( light blue ) molecules.

Question

Please help me understand it algebraically through conversion factors, I am not familiar with the logic

If the molecules in the above illustration react to form NH₃ according to the equation

N₂ + 3 H₂  2 NH₃

the limiting reagent is

the number of NH₃ molecules formed is

, and

molecules in excess is

Answer 1

Hello, John,

The question wants us to count the numbers of atoms we have of both N₂ and H₂ and use that information to answer the questions. A balanced equation tells us the number of individual atoms we need to complete a particular reaction. Do we have enough hydrogens to completely convert all of the nitrogens to NH₃, and so forth?

N₂ + 3 H₂  2 NH₃

In this case we need 3 H₂ molecules for every 1 N₂ molecule. I count the following:

molecules atoms

N₂ : 3 6

H₂ : 6 12

The molar ratio of H₂ to N₂ is 3 to 1, or 3. For a complete reaction, we would need three times as many hydrogen molecules/atoms as we have of nitrogen molecules/atoms. We don't have enough hydrogen to do that. Three molecules of N₂ would require 9 molecules of H₂, but we only have 6. That makes hydrogen the limiting reagent. Once it's gone, the reaction stops, and we'll have unreacted nitrogen remaining.

So let's work backwards and assume all of the hydrogen will be consumed. That means we can divide the hydrogen atoms by three to determine how many nitrogen atoms will be consumed. The 6 H atoms will form 2 ammonia molecules, NH₃. That means 2 of the N atoms are consumed, leaving 4 behind. Since nitrogen is a diatomic (as in hydrogen) we can write the 4 leftover N atoms as 2N₂. There are 2 molecules of N₂ in excess, in other words.

Bob

Answer 2

From the balanced equation, you can see that it takes 3 molecules of H₂ (light blue) to react with 1 molecule of N₂ (dark blue). Since you have 3 molecules of N₂, you will need 3 x 3 or 9 molecules of H₂.

Using conversion factors (dimensional analysis), this calculation would look like this....

3 molecules N₂ x 3 molecules H₂ / 1 molecule N₂ = 9 molecules of H₂ needed.

Limiting reagent is H₂ because you don't have enough (you have only 6 but you need 9).

Number of NH₃ molecules formed is dependent on the limiting reagent, as this will "limit" the amount you can make. Since we have only 6 molecules of H₂, the calculation looks like this...

6 molecules H₂ x 2 molecules NH₃ / 3 molecules H₂ = 4 molecules NH₃ formed

There will be excess N₂ left over. Since we have only 6 molecules H₂, there should be 1 molecule of N₂ left over. Calculation looks like this...

6 molecules H₂ x 1 molecule N₂ / 3 molecules H₂ = 2 molecules N₂ used up.

Started with 3 molecules N₂, so there is 1 molecule N₂ left over.