The reaction between nitrogen (N2) and hydrogen (H2) to form ammonia (NH3) is:
N2(g)+3H2(g)→2NH3(g)\mathrm{N}_2(g)+3\mathrm{H}_2(g)\rightarrow 2\mathrm{NH}_3(g)N2(g)+3H2(g)→2NH3(g)
According to the stoichiometry of the balanced equation, 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. If there is 1 mole of N2 and 1 mole of H2 initially present in a vessel:
- The reaction requires 3 moles of H2 for every 1 mole of N2 to fully react.
- Since only 1 mole of H2 is present, hydrogen is the limiting reactant.
- Nitrogen is in excess and will remain after the reaction stops.
At the molecular level, if a single N2 molecule reacts completely, it would require 3 H2 molecules. With 1 mole each of N2 and H2, only 1/3 mole of N2 can react with the 1 mole of H2. Thus, only 23\frac{2}{3}32 mole of NH3 can form from the reaction of 1 mole of H2 with 1 mole of N2 because the extent of reaction is limited by the availability of H2. If the reaction is allowed to proceed to equilibrium in a given volume, the equilibrium concentrations will depend on how much reaction occurs, but the initial limiting reactant scenario remains the same. The number of moles of NH3 formed will be based on the limiting H2 and the balanced stoichiometry.
