To determine which combination of molecules will produce ammonia with no leftovers, we need to analyze each option based on the stoichiometry of the reaction:
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The balanced equation: \[ 1 \text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3 \] This means for every 1 molecule of N₂, we need 3 molecules of H₂ to produce 2 molecules of NH₃.
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Option A: 2 N₂ and 6 H₂
- N₂: 2 molecules will produce \( 2 \times 2 = 4 \text{NH}_3 \).
- H₂: 6 molecules are used, which corresponds to \( 2 \times 3 = 6 \text{H}_2 \).
- This combination uses all of the reactants with none leftover.
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Option B: 8 N₂ and 4 H₂
- N₂: 8 molecules will produce \( 8 \times 2 = 16 \text{NH}_3 \).
- H₂: 4 molecules are needed, but for 8 N₂, we need \( 8 \times 3 = 24 \text{H}_2 \) which means there would be too few H₂ molecules.
- This combination would leave leftover N₂.
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Option C: 4 N₂ and 4 H₂
- N₂: 4 molecules will produce \( 4 \times 2 = 8 \text{NH}_3 \).
- H₂: 4 molecules can only provide enough for \( \frac{4}{3} \) N₂ (since 3 H₂ are needed for each N₂).
- Again, this combination leaves leftover N₂.
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Option D: 5 N₂ and 6 H₂
- N₂: 5 molecules will produce \( 5 \times 2 = 10 \text{NH}_3 \).
- H₂: 6 molecules of H₂ would only suffice for \( \frac{6}{3} = 2 \) N₂.
- So, this combination also leaves leftover N₂.
From this analysis, only Option A: 2 N₂ and 6 H₂ will produce ammonia with no leftovers.