Final answer:
At equilibrium, with 10 moles of N₂ and H₂ reacting and 40% of H₂ left, the total moles in the chamber are 16 moles, considering the stoichiometry of the reaction and the remaining reactants. The correct answer is (3) 16.
Step-by-step explanation:
The student asked what the total moles in the chamber would be if 10 moles of N2 and H2 are made to react, and at equilibrium, 40% of H2 was left. The balanced chemical reaction is N₂ + 3H₂ → 2NH₃.
If we assume that hydrogen is the limiting reagent, we start with 10 moles of H2 and at equilibrium, we have 40% left, which is 4 moles of H₂. Thus, 10 moles – 4 moles = 6 moles of H₂ reacted.
According to the 3:2 ratio from the balanced equation, if 3 moles of H2 give 2 moles of NH₃, then 6 moles of H₂ will give 4 moles of NH₃.
From the initial 10 moles of N₂, we use 2 moles (since it's reacting in a 1:3 ratio with H₂), which leaves us with 8 moles of N₂ unreacted.
Adding up the remaining H₂, the produced NH₃, and the unreacted N₂, we get a total of 4 + 4 + 8 = 16 moles in the chamber. The correct answer is (3) 16.