Nitrogen and hydrogen gases react to form ammonia gas as follows: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ At a certain temperature and pressure, \(1.2 \mathrm{~L}\) of \(\mathrm{N}_{2}\) reacts with \(3.6 \mathrm{~L}\) of \(\mathrm{H}_{2}\). If all the \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) are consumed, what volume of \(\mathrm{NH}_{3}\), at the same temperature and pressure, will be produced?

The balanced chemical equation for the reaction between nitrogen gas and hydrogen gas to form ammonia gas is given as: \[ \mathrm{N}_{2}(g) + 3\ \mathrm{H}_{2}(g) \longrightarrow 2 \ \mathrm{NH}_{3}(g) \]

Using the balanced chemical equation, we can determine the mole ratios of the reactants and products. In this case, the mole ratio of N2 to H2 is 1:3, and the mole ratio of N2 to NH3 is 1:2.

We are given the volumes of N2 and H2 reacting as 1.2 L and 3.6 L, respectively. Based on the mole ratios and using the balanced chemical equation, we can calculate the volume of NH3 produced as follows: First, we need to check which reactant is the limiting reactant. Since the mole ratio of N2 to H2 is 1:3, we can divide the volume of H2 by 3: \[ \frac{3.6 \ \mathrm{L}}{3} = 1.2 \ \mathrm{L} \] Since the volume of N2 is equal to this value, this means that both reactants are completely consumed in the reaction, and we can proceed with the calculation. Using the mole ratio of N2 to NH3 (1:2), we can now calculate the volume of NH3 produced: \[ 1.2 \ \mathrm{L} \times \frac{2 \ \mathrm{moles \ of \ NH}_{3}}{1 \ \mathrm{mole \ of \ N}_{2}} \] This simplifies to: \[ 1.2 \ \mathrm{L} \times 2 = 2.4 \ \mathrm{L} \]

Since all the N2 and H2 are consumed, the volume of NH3 produced, at the same temperature and pressure as the reactants, is 2.4 L.