Given 28 g of nitrogen and 9 g of hydrogen, which gas acts as the limiting reagent in the formation of ammonia?
Correct answer: B. H₂
- A. N₂
- B. H₂
- C. Both in equal proportion
- D. None
Explanation
Moles N₂ = 28/28 = 1 mol; Moles H₂ = 9/2 = 4.5 mol. According ot balanced chemical equation: N₂ + 3H₂ → 2NH₃. The reaction requires 3 mol H₂ per 1 mol N₂ → hydrogen is in excess, nitrogen is limiting.
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Mole calculations connect mass, number of particles and Avogadro's number, while balanced equations provide the mole ratios used in stoichiometry. Questions cover limiting and excess reactants, theoretical yield and percentage yield, including identifying which reactant is consumed first and comparing the actual product with the maximum possible product.
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