To produce ammonia (NH3) according to the given reaction:
\[ 1 \text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3 \]
Let's analyze each option to see which one provides the right combination of nitrogen (N2) and hydrogen (H2) that will yield ammonia without any leftovers.
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2N2 and 6H2:
- \(2 \text{N}_2\) will produce \(2 \times 2 = 4 \text{NH}_3\) (since 1 N2 produces 2 NH3).
- For \(2 \text{N}_2\), the required \(H_2\) is \(2 \times 3 = 6 \text{H}_2\) (since 1 N2 requires 3 H2).
- This combination produces ammonia with no leftovers.
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8N2 and 4H2:
- \(8 \text{N}_2\) will produce \(8 \times 2 = 16 \text{NH}_3\).
- For \(8 \text{N}_2\), the required \(H_2\) is \(8 \times 3 = 24 \text{H}_2\).
- There are not enough hydrogen molecules; hence, there would be leftovers.
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5N2 and 6H2:
- \(5 \text{N}_2\) will produce \(5 \times 2 = 10 \text{NH}_3\).
- For \(5 \text{N}_2\), the required \(H_2\) is \(5 \times 3 = 15 \text{H}_2\).
- Again, there are not enough hydrogen molecules, leading to leftovers.
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4N2 and 4H2:
- \(4 \text{N}_2\) will produce \(4 \times 2 = 8 \text{NH}_3\).
- For \(4 \text{N}_2\), the required \(H_2\) is \(4 \times 3 = 12 \text{H}_2\).
- There are not enough hydrogen molecules; hence, there would be leftovers.
The only combination that will produce ammonia with no leftovers is:
2N2 and 6H2.
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