Final answer:
Using the stoichiometry of the balanced chemical reaction N2 + 3H2 → 2NH3, 4.61 moles of N2 can produce 9.22 moles of NH3, which is approximately 156.98 grams of NH3 when using the molar mass of NH3 (17.03 g/mol).
Step-by-step explanation:
To find out how many grams of NH3 (ammonia) can be produced from 4.61 mol of N2 with excess H2 (hydrogen), we will use the stoichiometry of the balanced chemical equation N2 + 3H2 → 2NH3. First, we need to determine the mole ratio between N2 and NH3, which is 1:2. Since you have been given the reaction yield from hydrogen, for every 3 moles of H2, 2 moles of NH3 are produced, we can convert the amount of nitrogen to ammonia. This ratio from the balanced equation implies that for every 1 mole of N2, 2 moles of NH3 are produced.
Now, if you have 4.61 moles of N2, you would produce:
4.61 mol N2 × (2 mol NH3 / 1 mol N2) = 9.22 mol NH3
To convert moles of NH3 to grams, we'll use the molar mass of NH3, which is approximately 17.03 g/mol. Therefore, the mass of NH3 produced is:
9.22 mol NH3 × 17.03 g/mol = 156.98 g NH3
This calculation indicates that 4.61 moles of N2 can produce about 156.98 grams of NH3, assuming there is excess hydrogen and the reaction goes to completion.