How many grams of NH3 can be produced from the mixture of 3.00 g each of nitrogen and hydrogen by the Haber process? a) 2.00 g b) 3.00 g c) 3.65g d) 1.82 g
The Correct Answer and Explanation is:
To determine how many grams of ammonia (NH₃) can be produced from a mixture of 3.00 g of nitrogen (N₂) and 3.00 g of hydrogen (H₂) using the Haber process, we begin with the balanced chemical equation:
N₂ + 3H₂ → 2NH₃
Step 1: Calculate molar masses
- Molar mass of N₂ = 28.02 g/mol
- Molar mass of H₂ = 2.02 g/mol
- Molar mass of NH₃ = 17.03 g/mol
Step 2: Convert mass to moles
- Moles of N₂ = 3.00 g ÷ 28.02 g/mol ≈ 0.107 mol
- Moles of H₂ = 3.00 g ÷ 2.02 g/mol ≈ 1.485 mol
Step 3: Determine limiting reagent
From the balanced equation, 1 mol of N₂ reacts with 3 mol of H₂.
So, 0.107 mol of N₂ would need:
0.107 mol × 3 = 0.321 mol of H₂
We have 1.485 mol of H₂ available, which is more than enough.
Therefore, N₂ is the limiting reagent.
Step 4: Use limiting reagent to find moles of NH₃
From the equation, 1 mol of N₂ produces 2 mol of NH₃.
So, 0.107 mol of N₂ will produce:
0.107 mol × 2 = 0.214 mol of NH₃
Step 5: Convert moles of NH₃ to grams
0.214 mol × 17.03 g/mol ≈ 3.64 g
Final Answer:
The maximum mass of NH₃ that can be produced is approximately 3.65 g, so the correct option is:
c) 3.65 g
Explanation:
This problem demonstrates a classic limiting reagent scenario in stoichiometry. Even though equal masses of nitrogen and hydrogen are mixed, their different molar masses mean they contribute different numbers of moles. Since nitrogen molecules are heavier, fewer moles of nitrogen are present, and it becomes the limiting reagent. The reaction stops when nitrogen is used up, and the excess hydrogen remains unreacted. The amount of ammonia formed depends entirely on how much nitrogen reacts. By converting grams to moles, applying the balanced equation, and converting the product back to grams, we find the answer.
