Question

1.35 g H₂ is allowed to react with 10.2 g N₂, producing 1.25 g NH3.

Part A
What is the theoretical yield in grams for this reaction under the given conditions?
Express your answer to three significant figures and include the appropriate units.
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μÅ
?

Answer 1

The theoretical yield for the reaction is 12.38 g of NH₃.

Step-by-step explanation:

The given balanced equation for the reaction is:

N₂(g) + 3H₂(g) → 2NH₃(g)

To find the theoretical yield of NH₃, we need to calculate the amount of NH₃ that would be produced if all the reactants were completely converted to products.

The stoichiometry of the balanced equation tells us that 1 mole of N₂ produces 2 moles of NH₃.

So, let's calculate the number of moles of N₂:

Moles of N₂ = Mass of N₂ / Molar mass of N₂
Moles of N₂ = 10.2 g / 28.02 g/mol

= 0.364 mol

Since 1 mole of N₂ reacts to form 2 moles of NH₃, the theoretical yield of NH₃ would be:

Theoretical yield of NH₃ = 0.364 mol N₂ × (2 mol NH₃ / 1 mol N₂) × (17.03 g/mol NH₃)

= 12.38 g