In the formation reaction for NH3, a chemist begins with 50.0 g of hydrogen gas and an excess of nitrogen. How many grams of ammonia (NH3) will she be able to produce?

2 Answers

ncchemist's profile pic

ncchemist | eNotes Employee

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Ammonia (NH3) is an amine gas that is produced via an industrial chemical reaction called the Haber process.  It is shown below.

N2 + 3H2 --> 2NH3

During the Haber process, nitrogen gas (N2) and hydrogen gas (H2) are combined together under high temperature and pressure in the presence of a catalyst (usually a metal catalyst like iron) to produce ammonia gas (NH3).  The Haber process is incredibly important because although nitrogen gas makes up 80% of the Earth's atmosphere, it is very stable and chemically inert, thus making it difficult to convert into more chemically useful nitrogen containing substances.  The Haber process is responsible for the production of much of the nitrogen containing fertilizers in use today that allow modern agriculture to feed the world's population.

We see from the balanced chemical equation above that one mole of nitrogen combines with three moles of hydrogen to produce two moles of ammonia.  In your question, the chemist begins with 50 g of hydrogen gas as the limiting reagent (the nitrogen gas is in excess).  So first let's convert the hydrogen gas from grams to moles.

50 g H2 * (1 mole/2 g) = 25 moles H2

We know from the chemical equation that for every three moles of H2, two moles of NH3 are formed.  So now we can find the number of moles of NH3 that will be formed.

25 moles H2 * (2 moles NH3/3 moles H2) = 16.7 moles NH3

Now we can convert the moles of NH3 into grams using the molecular weight.

16.7 moles NH3 * (17 g/1 mole) = 283.9 g NH3

So the chemist will be able to produce 283.9 g of NH3.

ayl0124's profile pic

ayl0124 | Student, Grade 12 | (Level 1) Valedictorian

Posted on

First, write a balanced chemical equaion. This is crucial in making the correct ratios when you perform stoichiometry.

`"3H_2" + "N_2" -> "2NH_3"`

Both hydrogen and nitroge exist as diatomic gases in nature.

Now, simply use stoichiometry.

`50"gH_2" = (1"molH_2")/(2.016"gH_2") = (2"molNH_3")/(3"molH_2") = (17.034"gNH_3")/(1"molNH_3") = 281.646"gNH_3"`

With three sig figs, the final answer will be 282 g NH3.