Final answer:
To produce 2.17 mol of NH3, 30.42 grams of N2 are needed, based on the stoichiometry of the balanced chemical equation and the molar mass of N2.
Step-by-step explanation:
The student has asked to determine how many grams of N2 are needed to produce 2.17 mol of NH3. To answer the question, we need to understand the stoichiometry of the balanced chemical equation, which states that 1 mole of N2 gas reacts with 3 moles of H2 gas to form 2 moles of NH3 gas. To determine how many grams of N₂ are produced from the given reaction, we need to use the balanced chemical equation and the molar masses of the compounds involved.
From the balanced equation, we can see that 1 mole of N₂ reacts to produce 2 moles of NH₃. Using the molar masses of N₂ (28.02 g/mol) and NH₃ (34.08 g/mol), we can calculate the mass of N₂ required:
2.17 mol of NH₃ * (28.02 g N₂ / 2 mol NH₃) = 30.08 g N₂
Therefore, 30.08 grams of N₂ are needed to produce 2.17 moles of NH₃. The molar mass of N2 is 28.02 g/mol. Since 1 mole of N2 produces 2 moles of NH3, we need half the number of moles of N2 to produce NH3, which is 2.17 mol / 2 = 1.085 moles of N2. Multiplying this by the molar mass of N2 (1.085 mol × 28.02 g/mol) gives us 30.42 grams of N2 needed.