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The reaction between nitrogen and hydrogen to form ammonia is carried out in a flask containing 30.5 g of [tex]N_2[/tex] and 8.65 g of [tex]H_2[/tex]. What is the theoretical yield of [tex]NH_3[/tex] in grams?

\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]

A. 37.1 g [tex]NH_3[/tex]
B. 48.7 g [tex]NH_3[/tex]
C. 18.6 g [tex]NH_3[/tex]

Answer :

The theoretical yield of NH₃ is a) 37.1 grams.

This is calculated using stoichiometry and the molar masses of the reactants and product in the balanced equation. Since N₂ is the limiting reagent, it determines the amount of NH₃ produced.

To determine the theoretical yield of NH₃, we first need to use the balanced chemical equation for the reaction:

N₂(g) + 3H₂(g) → 2NH₃(g)

Next, we calculate the moles of N₂ and H₂ present:

  • Moles of N₂ = 30.5 g / 28.02 g/mol = 1.089 moles
  • Moles of H₂ = 8.65 g / 2.02 g/mol = 4.283 moles

The stoichiometry from the balanced equation shows that 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃. Now, let's find the limiting reagent:

  • 1.089 moles of N₂ would need 3 * 1.089 = 3.267 moles of H₂

Since there are 4.283 moles of H₂ available, which is more than the 3.267 moles needed, N₂ is the limiting reagent. We use the moles of the limiting reagent to determine the moles of NH₃ produced:

  • Moles of NH₃ = 1.089 moles of N₂ * (2 moles NH₃ / 1 mole N₂) = 2.178 moles NH₃

Finally, convert moles of NH₃ to grams:

  • Theoretical yield = 2.178 moles NH₃ * 17.03 g/mol = 37.1 grams of NH₃

complete question : The reaction between nitrogen and hydrogen to form ammonia is carried out in a flask containing 30.5 g of N₂​ and 8.65 g of H₂. What is the theoretical yield of NH₃ in grams?

N₂(g)+3H₂(g)→2NH₃​(g)

a 37.1 gNH₃

b ​48.7 gNH₃​

c 18.6 gNH₃​​

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