Approximately 4.30 x
atoms of ammonia would be created with the limiting reagent (nitrogen) in this reaction.
Based on the information provided, 10.71 moles of ammonia would be created with the limiting reagent. Here's the work:
Balanced chemical equation: The balanced chemical equation for the reaction is N₂ + 3H₂ → 2NH₃.
Moles of reactants: We are given that 100.0g of nitrogen and 100.0g of hydrogen are reacted. We can convert these masses to moles using their respective molar masses:
Moles of nitrogen (N₂) = 100.0g / 28.0g/mol = 3.57 mol
Moles of hydrogen (H₂) = 100.0g / 2.02g/mol = 49.50 mol
Limiting reactant: We need to determine which reactant is limiting. We can do this by comparing the moles of each reactant to the stoichiometric coefficients in the balanced equation.
According to the equation, 1 mol of N₂ requires 3 mol of H₂.
In this case, we have 3.57 mol of N₂ and 49.50 mol of H₂.
Dividing both sides of the inequality by 3.57 mol N₂, we get:
H₂/N₂ = 49.50 mol H₂ / 3.57 mol N₂ ≈ 13.86
Since 13.86 is greater than 3 (the stoichiometric ratio), hydrogen is present in excess, and nitrogen is the limiting reactant.
Moles of ammonia: Now that we know the limiting reactant, we can calculate the moles of ammonia produced. From the balanced equation, 1 mol of N₂ produces 2 mol of NH₃. Therefore:
Moles of NH₃ = 3.57 mol N₂ * (2 mol NH₃ / 1 mol N₂) = 7.14 mol NH₃
Atoms of ammonia: Finally, we can convert the moles of ammonia to atoms by multiplying by Avogadro's number (6.022 x
atoms/mol):
Atoms of NH₃ = 7.14 mol NH₃ * 6.022 x
atoms/mol ≈ 4.30 x
atoms
Approximately 4.30 x
atoms of ammonia would be created with the limiting reagent (nitrogen) in this reaction.