To determine which combination of molecules will produce ammonia with no leftovers, we can analyze the stoichiometry of the reaction:
The reaction is: \[ 1N_2 + 3H_2 \rightarrow 2NH_3 \]
From this equation, we can see that:
- 1 molecule of \(N_2\) reacts with 3 molecules of \(H_2\) to produce 2 molecules of \(NH_3\).
Now, let's evaluate each option:
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5N2 and 6H2:
- For 5 molecules of \(N_2\), you would need \(5 \times 3 = 15\) molecules of \(H_2\).
- This combination will produce ammonia, but it has leftover \(H_2\) (15 needed, 6 provided).
-
2N2 and 6H2:
- For 2 molecules of \(N_2\), you would need \(2 \times 3 = 6\) molecules of \(H_2\).
- This combination will produce \(2 \times 2 = 4\) molecules of \(NH_3\) and uses both \(N_2\) and \(H_2\) with no leftovers.
-
8N2 and 4H2:
- For 8 molecules of \(N_2\), you would need \(8 \times 3 = 24\) molecules of \(H_2\).
- This combination will produce ammonia, but it has leftover \(N_2\) (24 needed, 4 provided).
-
4N2 and 4H2:
- For 4 molecules of \(N_2\), you would need \(4 \times 3 = 12\) molecules of \(H_2\).
- This combination will produce ammonia, but it has leftover \(H_2\) (12 needed, 4 provided).
Based on this analysis, the correct answer is:
2N2 and 6H2.
This combination produces ammonia with no leftovers.