To determine which combination of molecules will produce ammonia (NH3) with no leftovers, we can analyze each option based on the stoichiometry of the reaction:
The balanced equation shows that:
- 1 molecule of N2 reacts with 3 molecules of H2 to produce 2 molecules of NH3.
We can express the proportions needed for the production of NH3:
- For every 1 N2, we need 3 H2. Therefore, the ratio can be expressed as: \[ \text{N}_2 : \text{H}_2 = 1 : 3 \] or equivalently: \[ \text{N}_2 : \text{NH}_3 = 1 : 2 \quad \text{and} \quad \text{H}_2 : \text{NH}_3 = 3 : 2 \]
Now, let's evaluate the options one by one to see which one will yield ammonia with no leftovers:
-
8N2 and 4H2:
- Molecules of NH3 produced = 2 × (8 N2) = 16 NH3
- Molecules of H2 needed = (3 × 8) = 24 H2
- Leftover: 4 H2 (24 needed - 4 available = 20 leftover)
-
2N2 and 6H2:
- Molecules of NH3 produced = 2 × (2 N2) = 4 NH3
- Molecules of H2 needed = (3 × 2) = 6 H2
- Leftover: 0 H2 (6 needed - 6 available = 0 leftover)
-
4N2 and 4H2:
- Molecules of NH3 produced = 2 × (4 N2) = 8 NH3
- Molecules of H2 needed = (3 × 4) = 12 H2
- Leftover: 8 H2 (12 needed - 4 available = 8 leftover)
-
5N2 and 6H2:
- Molecules of NH3 produced = 2 × (5 N2) = 10 NH3
- Molecules of H2 needed = (3 × 5) = 15 H2
- Leftover: 9 H2 (15 needed - 6 available = 9 leftover)
After analyzing these options, the only combination that produces ammonia with no leftovers is:
2N2 and 6H2.
That is the correct answer.