The maximum mass of NH₃ that can be formed when 36.52 g of N₂ reacts with 10.62 g of H₂ is 44.35 g
<h3>Balanced equation </h3>
N₂ + 3H₂ —> 2NH₃
Molar mass of N₂ = 14 × 2 = 28 g/mol
Mass of N₂ from the balanced equation = 1 × 28 = 28 g
Molar mass of H₂ = 2 × 1 = 2 g/mol
Mass of H₂ from the balanced equation = 3 × 2 = 6 g
Molar mass of NH₃ = 14 + (3×1) = 17 g/mol
Mass of NH₃ from the balanced equation = 2 × 17 = 34 g
SUMMARY
From the balanced equation above,
28 g of N₂ reacted with 6 g of H₂ to produce 34 g of NH₃
<h3>How to determine the limiting reactant </h3>
From the balanced equation above,
28 g of N₂ reacted with 6 g of H₂
Therefore,
36.52 g of N₂ will react with = (36.52 × 6) / 28 = 7.83 g of H₂
From the above calculation, we can see that only 7.83 g out of 10.62 g of H₂ are required to react completely with 36.52 g of N₂.
Therefore, N₂ is the limiting reactant
<h3>How to determine the maximum mass of NH₃ produced </h3>
From the balanced equation above,
28 g of N₂ reacted to produce 34 g of NH₃
Therefore,
36.52 g of N₂ will react to produce = (36.52 × 34) / 28 = 44.35 g of NH₃
Thus, the maximum mass of NH₃ obtained from the reaction is 44.35 g
Learn more about stoichiometry:
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