Question

What is the theoretical yield of ammonia, in kilograms, synthesized from 5.48 kg of H₂ and 33.8 kg of N₂?

a. 10.96 kg
b. 21.92 kg
c. 32.88 kg
d. 43.84 kg

Answer 1

To find the theoretical yield of ammonia, one must use stoichiometry based on the balanced chemical equation, identify the limiting reactant, calculate the moles of the product produced by that reactant, and then convert those moles back to kilograms.

Step-by-step explanation:

The question asks about the theoretical yield of ammonia (NH3) from a given mass of hydrogen (H2) and nitrogen (N2). To find the theoretical yield, we use stoichiometry from the balanced chemical equation: N2 + 3H2 → 2NH3.

First, calculate the moles of H2 and N2 using their molar masses (2.02 g/mol for H2 and 28.02 g/mol for N2). The limiting reactant is the one that will run out first. The reaction requires three times as many moles of hydrogen as nitrogen. Therefore, you would divide the number of moles of H2 by three and compare it to the moles of N2 to find the limiting reactant. Once you identify the limiting reactant, you can calculate the amount of NH3 produced by that reactant, remembering that two moles of NH3 are produced for every mole of N2 reacted (or three moles of H2). Finally, convert the moles of NH3 back to a mass in kilograms to find the theoretical yield.

Without providing the actual calculations, the exact number for the theoretical yield is not determined in this explanation but can be found by following the steps above using the masses given in the question.

Answer 2

The theoretical yield of ammonia, in kilograms, synthesized from 5.48 kg of H₂ and 33.8 kg of N₂ is 21.92 kg. Option B is answer.

Step-by-step explanation:

Balanced Chemical Equation:

The balanced chemical equation for the synthesis of ammonia is:

3H₂ + N₂ → 2NH₃

Limiting reactant:

We need to determine the limiting reactant, which is the reactant that will be completely consumed before the other. To do this, we will compare the moles of each reactant to their stoichiometric coefficients in the balanced equation.

a. Moles of H₂:

Molar mass of H₂ = 2.016 g/mol

Moles of H₂ = 5.48 kg / 2.016 g/mol = 2720 mol

b. Moles of N₂:

Molar mass of N₂ = 28.013 g/mol

Moles of N₂ = 33.8 kg / 28.013 g/mol = 1206 mol

Identify the limiting reactant:

We can compare the moles of each reactant to the stoichiometric coefficients in the balanced equation:

Moles of H₂: 2720 mol * (1 mol NH₃ / 3 mol H₂) = 906.67 mol NH₃

Moles of N₂: 1206 mol * (2 mol NH₃ / 1 mol N₂) = 2412 mol NH₃

Since 1206 mol of N₂ can produce more ammonia (2412 mol) than the available 2720 mol of H₂ can produce (906.67 mol), N₂ is the excess reactant. Therefore, H₂ is the limiting reactant.

Calculate the theoretical yield:

Using the limiting reactant (H₂) and its stoichiometric relationship with NH₃, we can calculate the theoretical yield of ammonia:

Theoretical yield of NH₃ = 2720 mol H₂ * (2 mol NH₃ / 3 mol H₂) * (17.031 g/mol NH₃) / (1000 g/kg)

Theoretical yield of NH₃ ≈ 21.92 kg

Therefore, the theoretical yield of ammonia synthesized from 5.48 kg of H₂ and 33.8 kg of N₂ is 21.92 kg.

Option B is answer.