First, let's write the balanced chemical equation for the reaction between N2 and H2 to form NH3:
$$N_2 + 3H_2 \rightarrow 2NH_3$$
Next, we need to determine the limiting reactant. To do this, we need to compare the number of moles of each reactant.
At STP (0 °C and 1 atm), 1 liter of N2 contains (1 L / 22.4 L/mol) = 0.0446 mol of N2, and 1 liter of H2 contains (1 L / 22.4 L/mol) = 0.0446 mol of H2.
Based on the balanced chemical equation, 1 mole of N2 reacts with 3 moles of H2. Therefore, we need 0.0446 mol * 3 = 0.1338 mol of H2 to react completely with 0.0446 mol of N2.
Since we only have 0.0446 mol of H2, it is the limiting reactant.
Now, we can use the stoichiometry of the balanced chemical equation to determine how many moles of NH3 will be produced.
Since 1 mole of H2 reacts to produce 2 moles of NH3, 0.0446 mol of H2 will produce 0.0446 mol * 2 = 0.0892 mol of NH3.
Finally, we can use the ideal gas law to determine the volume of NH3 produced. At STP, 1 mole of any gas occupies 22.4 liters. Therefore, 0.0892 mol of NH3 will occupy 0.0892 mol * 22.4 L/mol = 1.99 liters.
Therefore, 2 liters of N2 and H2 on complete reaction would give 1.99 liters of NH3.
