Question

A mixture of N₂ (g) and H₂ (g) reacts in a closed container to form ammonia, NH₃ (g). At this stage, 1.8 mol N₂, 1.8 mol H₂, and 1.8 mol NH₃ are present. How many moles of N₂ and H₂ were present originally?

A. 1.8 mol N₂ and 0 mol H₂
B. 0 mol N₂ and 1.8 mol H₂
C. 0.9 mol N₂ and 0.9 mol H₂
D. 1.8 mol N₂ and 1.8 mol H₂
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Answer 1

Based on the stoichiometry of the reaction N₂(g) + 3 H₂(g) ⇒ 2 NH₃(g), we deduce that to form the 1.8 moles of NH₃ present, we would have started with 1.8 moles of N₂ and 2.7 moles of H₂.

Step-by-step explanation:

To determine the original amounts of N₂ (g) and H₂ (g) present before the reaction, we have to consider the stoichiometry of the reaction: N₂(g) + 3 H₂(g) ⇒ 2 NH₃(g). According to the balanced equation, one mole of N₂ will react with three moles of H₂ to produce two moles of NH₃. The reaction alsostates that mass and the number of atoms must be conserved. Since we currently have 1.8 moles of NH₃, we can work backwards to find the original moles of reactants.

From the stoichiometry of the balanced chemical equation, we deduce that:

• 1 mole of N₂ produces 2 moles of NH₃, therefore 1.8 moles of NH₃ would come from 0.9 moles of N₂.
• 3 moles of H₂ react to produce 2 moles of NH₃, therefore to produce 1.8 moles of NH₃, we would need 1.8 * (3/2) = 2.7 moles of H₂.

Since we started with 1.8 moles of N₂ and have 0.9 moles in the reaction mixture, no N₂ has been consumed. However, we started with an unknown amount of H₂ and have only 1.8 moles left. By subtracting 1.8 moles of H₂ present from 2.7 moles needed to form the NH₃, we find that 0.9 moles of H₂ has been consumed. We can then infer that we originally had 1.8 moles + 0.9 moles = 2.7 moles of H₂.

Therefore, we started with 1.8 moles of N₂ and 2.7 moles of H₂.