Questions: Consider the Haber-Bosch process for the synthesis of ammonia from its elements. Calculate the theoretical yield in moles NH3 from the complete reaction of 59.8 grams N2 in the presence of excess H2 gas according to the following balanced chemical equation: N2(g) + 3 H2(g) -> 2 NH3(g)

Consider the Haber-Bosch process for the synthesis of ammonia from its elements. Calculate the theoretical yield in moles NH3 from the complete reaction of 59.8 grams N2 in the presence of excess H2 gas according to the following balanced chemical equation: N2(g) + 3 H2(g) -> 2 NH3(g)
Transcript text: Consider the Haber-Bosch process for the synthesis of ammonia from its elements. Calculate the theoretical yield in moles $\mathrm{NH}_{3}$ from the complete reaction of 59.8 grams $\mathrm{N}_{2}$ in the presence of excess $\mathrm{H}_{2}$ gas according to the following balanced chemical equation: \[ \mathrm{N}_{2}(\mathrm{~g})+3 \quad \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \quad \mathrm{NH}_{3}(\mathrm{~g}) \]
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Solution

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Solution Steps

Step 1: Convert grams of N₂ to moles of N₂

Molar mass of N₂ = 28.02 g/mol

Moles of N₂ = (59.8 g N₂) * (1 mol N₂ / 28.02 g N₂) = 2.13 mol N₂

Step 2: Convert moles of N₂ to moles of NH₃ using the stoichiometric ratio

From the balanced chemical equation: N₂(g) + 3H₂(g) → 2NH₃(g) We can see that 1 mol of N₂ reacts to produce 2 moles of NH₃.

Moles of NH₃ = (2.13 mol N₂) * (2 mol NH₃ / 1 mol N₂) = 4.27 mol NH₃

Final Answer:

The theoretical yield of NH₃ is 4.27 moles.

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