To determine which combination of molecules will produce ammonia with no leftovers, we need to analyze the stoichiometry of the reaction:
The balanced equation is: \[ 1N_2 + 3H_2 \rightarrow 2NH_3 \]
From the equation, we can see that:
- 1 molecule of nitrogen (N₂) reacts with 3 molecules of hydrogen (H₂) to produce 2 molecules of ammonia (NH₃).
Now let's evaluate each option:
-
4N₂ and 4H₂:
- From 4N₂, we can produce \(4 \times 2 = 8NH₃\).
- This requires \(4 \times 3 = 12H₂\).
- Not enough hydrogen, leftovers of N₂.
-
2N₂ and 6H₂:
- From 2N₂, we can produce \(2 \times 2 = 4NH₃\).
- This requires \(2 \times 3 = 6H₂\).
- Perfectly used up, no leftovers.
-
5N₂ and 6H₂:
- From 5N₂, we can produce \(5 \times 2 = 10NH₃\).
- This requires \(5 \times 3 = 15H₂\).
- Not enough hydrogen, leftovers of N₂.
-
8N₂ and 4H₂:
- From 8N₂, we can produce \(8 \times 2 = 16NH₃\).
- This requires \(8 \times 3 = 24H₂\).
- Not enough hydrogen, leftovers of N₂.
The only combination that produces ammonia with no leftovers is:
2N₂ and 6H₂.