Final answer:
The amount of NH₃ produced from the reaction of 9.5 g of N₂ with 5.7 g of H₂ is calculated by first finding the limiting reactant by converting the mass of each to moles. It's found that N₂ is limiting, leading to 0.678 moles of NH₃ which translates to 11.54 g of NH₃, though this result doesn't match the given options, suggesting a discrepancy.
Step-by-step explanation:
To determine the amount of NH₃ produced from the reaction of 9.5 g of N₂ with 5.7 g of H₂, we first need to calculate the number of moles of each reactant. We can use the molar mass of nitrogen (28.02 g/mol) and hydrogen (2.02 g/mol) to convert the mass of N₂ and H₂ to moles.
- Moles of N₂ = 9.5 g / 28.02 g/mol = 0.339 moles
- Moles of H₂ = 5.7 g / 2.02 g/mol = 2.822 moles
Using the balanced equation N₂ + 3H₂ → 2NH₃, we can see that 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃. Therefore, nitrogen is the limiting reactant since we do not have enough H₂ to react with all of the N₂ (0.339 moles N₂ would require 1.017 moles H₂ but we only have 0.339 moles of H₂).
Now, we can find out how much NH₃ will be produced from the N₂:
- Moles of NH₃ produced = 0.339 moles N₂ * (2 moles NH₃ / 1 mole N₂) = 0.678 moles NH₃
- Mass of NH₃ produced = 0.678 moles * 17.03 g/mol (molar mass of NH₃) = 11.54 g
Be aware that the mass we obtained exceeds the options provided, suggesting a possible error in our assessment or the options given. We rounded our figures for this calculation, but with exact molar masses and more precise calculations, the result would be slightly different and may match one of the option choices.