3 moles N2(g) is allowed to react with 6 moles of H2(g) to formNH3(g) in a close container. Find excess reagent and remainingmoles of excess reagent?
Question
3 moles N2(g) is allowed to react with 6 moles of H2(g) to formNH3(g) in a close container. Find excess reagent and remainingmoles of excess reagent?
Solution
The balanced chemical equation for the reaction is:
N2(g) + 3H2(g) -> 2NH3(g)
From the balanced equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to form 2 moles of NH3.
Given that we have 3 moles of N2 and 6 moles of H2, we can see that the ratio of N2 to H2 is 1:2, which is less than the required ratio of 1:3 for the reaction to proceed completely.
Therefore, N2 is the limiting reagent and H2 is the excess reagent.
To find the remaining moles of the excess reagent, we need to calculate how much H2 is used up by the 3 moles of N2.
From the balanced equation, we know that 1 mole of N2 reacts with 3 moles of H2. Therefore, 3 moles of N2 will react with 3*3 = 9 moles of H2.
However, we only have 6 moles of H2. Therefore, all 6 moles of H2 will be used up and there will be no remaining moles of the excess reagent.
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