A mixture of \(\mathrm{N}_{2}(g)\) and \(\mathrm{H}_{2}(g)\) reacts in a closed container to form ammonia, \(\mathrm{NH}_{3}(g)\). The reaction ceases before either reactant has been totally consumed. At this stage \(3.0 \mathrm{~mol} \mathrm{~N}_{2}, 3.0\) \(\mathrm{mol} \mathrm{H}_{2},\) and \(3.0 \mathrm{~mol} \mathrm{NH}_{3}\) are present. How many moles of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) were present originally?

The balanced equation for the reaction is: \[N_2(g) + 3H_2(g) \rightarrow 2NH_3(g).\]

We know that after the reaction has ceased, there is: - 3.0 moles of N2 - 3.0 moles of H2 - 3.0 moles of NH3

Let's assume that x moles of N2 have reacted and y moles of H2 have reacted. From the balanced equation, we know that: - 1 mole of N2 reacts with 3 moles of H2 to form 2 moles of NH3 - x moles of N2 will react with 3x moles of H2 to form 2x moles of NH3 - Now we can sum the moles of each reactant and product, according to the given data: - N2: 3.0 (final) = x (reacted) + original (unknown) - H2: 3.0 (final) = 3x (reacted) + original (unknown) - NH3: 3.0 (final) = 2x (reacted)

From the third equation, we find the amount of NH3 formed by the reaction as 3/2 times the moles of N2 reacted: - 3.0 (final) = 2x (reacted) - x = 1.5 moles

Now let's plug in the value of x into the equations for N2 and H2: - N2: 3.0 (final) = 1.5 (reacted) + original N2 - H2: 3.0 (final) = 3(1.5) (reacted) + original H2

Solving for the original moles of N2 and H2: - Original N2 = 3.0 (final) - 1.5 (reacted) = 1.5 moles - Original H2 = 3.0 (final) - 4.5 (reacted) = -1.5 moles However, a negative amount of moles is impossible. Recall that the reaction ceased before either reactant was consumed completely. This indicates that the amount of H2 calculated is incorrect due to the reaction not being carried out to completion. So, we reevaluate the stoichiometry based on the limiting reactant, H2: If the entire 3 moles of H2 were present initially and completely reacted, then 1 mole of H2 will react with 1/3 mole of N2, and therefore 3 moles of H2 will react with 1 mole of N2. This would result in: - Original N2: 3.0 (final) - 1.0 (reacted) = 2.0 moles - Original H2: 3.0 (final) - 0 (reacted) = 3.0 moles Thus, the correct answer is that there were originally 2.0 moles of N2 and 3.0 moles of H2 present.