theoretical yield of ammonia (NH₃) = 121.38 g
The limiting reactant is nitrogen (N₂) and the excess reactant is hydrogen (H₂).
Explanation:
We have the following chemical reaction in which nitrogen react with hydrogen to produce ammonia:
N₂ + 3 H₂ → 2 NH₃
Now we need to calculate the number of moles of each reactant:
number of moles = mass / molar weight
number of moles of N₂ = 100 / 28 = 3.57 moles
number of moles of H₂ = 100 / 2 = 50 moles
We see from the chemical reaction that 3 moles of H₂ react with 1 mole of N₂ so 50 moles of H₂ react with 16.67 moles of N₂ which is way more than the available N₂ quantity of 3.57 moles, so the limiting reactant is nitrogen (N₂) and the excess reactant is hydrogen (H₂).
Knowing this we devise the following reasoning:
if 1 mole of N₂ produces 2 moles of NH₃
then 3.57 moles of N₂ produces X moles of NH₃
X = (3.57 × 2) / 1 = 7.14 moles of NH₃
mass = number of moles × molar weight
mass of NH₃ = 7.14 × 17 = 121.38 g (theoretical yield)
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limiting reactant
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