Answer:
- 34 g of NH₃ will be produced from the reaction of 28 g of N₂ with 25 g of H₂.
Step-by-step explanation:
1) Balanced chemical equation
- N₂ (g) + 3H₂ (g) → 2NH₃(g)
2) Stoichiometric (theoretical ) mole ratios
- 1 mol N₂ (g) : 3mol H₂ (g) : 2 mol NH₃(g)
3) Number of moles of each reactant
- number of moles = mass in grams / atomic mass
- number of moles of N₂ = 28 g / 28 g/mol = 1 mol
- number of moles of H₂: 25 g / 2 g/mol = 12.5 mol
4) Limiting reactant
Since the stoichiometry states that 1 mol of N₂ reacts with 3 moles of H₂, the given mass of N₂ will react completely with the given amount of H₂, and the calculations must be done with the 28 g (1 mol) of N₂ as the limiting reactant.
5) Yield
Set the proportion with the mole ratios:
1 mol H₂ / 2 mol NH₃ = 1 mol H₂ / x ⇒ x = 2 mol NH₃
6) Convert to grams
- mass in grams = number of moles × molar mass = 2 mol × 17 g/mol = 34 g.
Answer: the reaction of 28 g of N₂ with 25 g of H₂ will produce 34 g of NH₃