To determine which combination of molecules will produce ammonia with no leftovers, we can look at the stoichiometry of the reaction:
\[ \text{1} \text{N}_2 + \text{3} \text{H}_2 \rightarrow \text{2} \text{NH}_3 \]
This means that:
- 1 molecule of N₂ reacts with 3 molecules of H₂ to produce 2 molecules of NH₃.
To analyze the provided options, we need to check each combination to see if they adhere to this stoichiometric balance:
-
4N₂ and 4H₂:
- N₂: 4
- H₂ required: \( 4 \times 3 = 12 \)
- Leftover H₂: \( 4 - 12 = -8 \) (deficit)
-
2N₂ and 6H₂:
- N₂: 2
- H₂ required: \( 2 \times 3 = 6 \)
- Produces \( 2 \times 2 = 4 \) NH₃ with no leftovers. This combination works perfectly.
-
8N₂ and 4H₂:
- N₂: 8
- H₂ required: \( 8 \times 3 = 24 \)
- Leftover H₂: \( 4 - 24 = -20 \) (deficit)
-
5N₂ and 6H₂:
- N₂: 5
- H₂ required: \( 5 \times 3 = 15 \)
- Leftover H₂: \( 6 - 15 = -9 \) (deficit)
From the calculations above, the only combination that produces ammonia without leftovers is:
2N₂ and 6H₂.