How Much Nitrogen and Hydrogen Are Needed to Form 6 Moles of Ammonia

The production of ammonia is a crucial industrial process, widely utilized in the production of fertilizers. Understanding the chemical reaction involved and the quantitative relationships between nitrogen, hydrogen, and ammonia is essential for optimizing this process.

Chemical Reaction and Stoichiometry

The balanced equation for the formation of ammonia (NH3) from nitrogen (N2) and hydrogen (H2) is:

N2 3 H2 → 2 NH3

This equation tells us that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. To form 6 moles of ammonia, we need to calculate the amounts of nitrogen and hydrogen required, and their respective masses.

Calculating Nitrogen Needed

Using the balanced equation, the mole ratio between N2 and NH3 is 1:2. Therefore, for every 2 moles of NH3, 1 mole of N2 is needed. To form 6 moles of NH3, we need:

6 moles NH3 x (1 mole N2 / 2 moles NH3) 3 moles N2

Calculating the Mass of Nitrogen

The molar mass of N2 is calculated as follows:

MN2 2 x 14.007 g/mol 28.014 g/mol

Therefore, the mass of 3 moles of N2 is:

mN2 n x MN2 3 mol x 28.014 g/mol 84.042 g N2

Conclusion

To form 6 moles of ammonia, 84.042 grams of nitrogen (N2) are required, assuming an excess of hydrogen and a pressure of 200 atm and temperature of 450°C in the presence of an iron catalyst, as in the Haber process.

Additional Information on Ammonia Formation

The production of ammonia is an exothermic reaction, meaning it releases heat. In the industrial process known as the Haber process, N2 and H2 react at high pressure and temperature (200 atm, 450°C) in the presence of an iron catalyst to produce NH3.

Stoichiometric Relationship Between Gases

The balanced equation for this reaction is:

N2 3 H2 → 2 NH3

This equation shows a 1:3:2 mole ratio between N2, H2, and NH3.

Calculations for Hydrogen Required

To form 6 moles of NH3, the moles of H2 required can be calculated using the mole ratio from the balanced equation:

6 mol NH3 x (3 mol H2 / 2 mol NH3) 9 mol H2

Note that this calculation is based on the assumption that hydrogen is in excess, and nitrogen is the limiting reactant.

Molar Mass and Mass of Hydrogen

The molar mass of H2 is:

MH2 2 x 1.008 g/mol 2.016 g/mol

Therefore, the mass of 9 moles of H2 is:

mH2 n x MH2 9 mol x 2.016 g/mol 18.144 g H2