Final answer:
To calculate the mass of NH₃ formed from 14g of N₂ with excess H₂, one must use stoichiometry to determine the moles of N₂ involved, identify the molar ratio from the balanced equation, determine the limiting reactant (which is N₂), and finally convert the moles of NH₃ produced to mass, which would be 17.03 grams.
Step-by-step explanation:
When 14g of N₂ are combined with excess H₂, the mass of NH₃ formed can be calculated using stoichiometry. According to the balanced chemical equation, N₂(g) + 3H₂(g) → 2NH₃(g), one mole of N₂ reacts with three moles of H₂ to produce two moles of NH₃.
To find the mass of NH₃ formed:
First, calculate the number of moles of N₂ using its molecular weight (14g / 28.02g/mol = 0.5 moles).
Then, using the molar ratio from the balanced equation (1:2 for N₂ to NH₃), calculate the moles of NH₃ that would form (0.5 moles N₂ x 2 moles NH₃/1 mole N₂ = 1 mole NH₃).
Finally, convert moles of NH₃ to mass (1 mole NH₃ x 17.03g/mol = 17.03 grams).
Since the equation states excess H₂, N₂ is the limiting reactant, and all 14g will be used to form NH₃.
The Ideal Gas Law is not needed in this situation because we are dealing with definite quantities and ratios, and not trying to find out pressure, volume, or temperature conditions of gases.