Question

In the production of ammonia, N2 + 3H2 --> 2NH3, if you start with 3 mol of N2 and 6 mol of H2, identify the limiting reagent and the excess reagent.

Answer 1

From the given chemical reaction, we determine the amount of NH3 formed from the amounts of the reactants.

N2: (3 mol N2) x (2 mols NH3 / 1 mole N2) = 6 moles NH3
H2: (6 mol H2) x (2 mols NH3 / 3 moles H2) = 4 moles NH3
Since there are fewer of the NH3 that can be formed from the H2. Then, it is the limiting reagent and the N2 is the excess reagent.