Given that you have 14.5 moles of N2, how many moles of H2 are need to produce 22.5 moles of NH3 according to the reaction: N2+ 3 H2 yields 2 NH3?

I am having extreme difficulty with this, I don't need the specific answer, just an explanation for this would be great please!

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The formula for the reaction is:

N2 + 3H2 --> 2 NH3

It is clear from this formula that 1 molecule of N2 combines with 3 molecules of H2 to produce 2 molecules of NH3. Therefore the the quantities of N2, H2 and NH3 involved in this reaction in terms of moles will also be in the same proportion. That is 1 mole of N2 will combine with 3 molecules of H2 to produce 2 molecules of NH3.

It is given that quantity of N2 used in the reaction is 14.5 moles.

Therefore quantity of H2 required will be 3 times this. Thus:

Quantity of H2 in moles = 14.5 x 3 = 43.5 moles.

Similarly quantity of NH3 produced = 14.5 x 2 = 29 moles.

Explanation:

The ratio of quantities of different chemical involved in a chemical reactions in terms of number of molecules and number of moles is same because each mole contains the same number of molecules. This constant number of molecules in a mole of substance is called *Avogadro Constant*, and is equal to 602,213,670 x 10^15.

Please note that the the data given in the question that 14.5 moles of N2 will combine with H2 in a chemical reaction leading to their conversion into 22.5 moles of NH3, is not valid.

There would be a excess of about 3 mol according to the reaction taking place.

Given molecular formula of the chemical equation:

N2+3H2 = 2NH3. It implies that one mole of N2 and 3moles of H2 by parts comine completely resulting in the product of 2 moles of NH3. The question is, then, in the same proportion, how many moles of H2 by part produce in 22.5 moles of NH3. Therefore, by ratio and proportion we can write this in the following form.

Moles of H2):moles of (H2 ) : moles (of NH3)

1 : 3 : 2

x /3 : x : 22.5.

So , x/3 = x/3 = 22.5/2 .

x = 3*22.5/2 = 33.75 moles of H2 are required to get 22.5 moles of NH3.

So, the product of 22.5 moles of NH3 requires, 33.75 moles of N2 and 33.75/3 =11.25 moles of H2. Therefore, out of the given 14.5 moles of H2, only 11.25 moles of H2 are sufficient to combine with 33.75 moles of N2 to get22.5 moles of NH3. There remains, thus, an excess of (14.5-11.25) moles = 3.25 moles of H2.

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