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Anonymous asked in Science & MathematicsChemistry · 1 month ago

Chemistry formulas?

theoretical and yield

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  • 1 month ago

    Molar mass of N₂ = 14.0 × 2 g/mol = 28.0 g/mol

    Molar mass of NH₃ = (14.0 + 1.0×3) g/mol = 17.0 g/mol

    Balanced equation for the reaction:

    N₂(g) + 3H₂(g) → 2NH₃(g)

    Mole ratio N₂ : NH₃ = 1 : 2

    Moles of N₂ reacted = (5.01 g) / (28.0 g/mol) = 0.1789 mol

    Maximum moles of NH₃ produced = (0.1789 mol) × 2 = 0.3578 mol

    Theoretical yield of NH₃ = (0.3578 mol) × (17.0 g/mol) = 6.08 g

    Percent yield = (4.20/6.08) × 100% = 69.1%

    ====

    OR:

    (5.01 g N₂) × (1 mol N₂ / 28.0 g N₂) × (2 mol NH₃ / 1 mol N₂) × (17.0 g NH₃ / 1 mol NH₃)

    = 6.08 g NH₃ (Theoretical yield)

    (4.20/6.08) × 100%

    = 69.1% yield

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