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Common Graded Assignment - CHEM 107

1. Write a balanced thermochemical equation with phase labels for the Haber process with the heat energy as part of the equation. (3 pts)



2. What is the theoretical yield of ammonia (in grams) if 16.55 grams of nitrogen gas and 10.15 grams of hydrogen gas are allowed to react? (9 pts)



3. Based on your theoretical yield, what is the percent yield of ammonia if only 8.33 grams of ammonia is produced? (3 pts)



4. How much heat energy (in kJ) will be absorbed or released if 8.33 grams of ammonia is produced? State whether the energy will be absorbed or released. (4 pts)



5. Ammonia gas reacts with aqueous nitric acid to form aqueous ammonium nitrate, a fertilizer.

(a) Write the balanced chemical equation for the reaction, including the physical states. (2 pts)

(b) Write the complete ionic and the net ionic equations for the reaction. (7 pts)

User Sniffer
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1 Answer

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The balanced thermochemical equation for the Haber process yields a theoretical ammonia yield of 20.06 g with a percent yield of 41.5%. The subsequent sections outline heat calculations and the net ionic equation for ammonia reacting with nitric acid.

The balanced thermochemical equation for the Haber process is as follows:

Nitrogen gas (N₂) + 3 Hydrogen gas (H₂) → 2 Ammonia (NH₃), ΔH = -92.4 kJ

This equation indicates that one mole of nitrogen gas (N₂) reacts with three moles of hydrogen gas (H₂) to produce two moles of ammonia (NH₃). The reaction is exothermic, releasing 92.4 kJ of heat energy.

To find the theoretical yield of ammonia, we first need to determine the limiting reactant. The balanced equation shows a 1:3 ratio between nitrogen gas and hydrogen gas. Let's calculate the moles of each reactant:

Moles of N₂ = mass / molar mass = 16.55 g / 28.02 g/mol ≈ 0.590 mol

Moles of H₂ = mass / molar mass = 10.15 g / 2.016 g/mol ≈ 5.04 mol

The ratio of moles of N₂ to H₂ is approximately 1:3, indicating that N₂ is the limiting reactant.

Now, using the stoichiometry of the reaction, we can determine the moles of NH₃ formed:

Moles of NH₃ = (moles of N₂) × (2 mol NH₃ / 1 mol N₂) ≈ 0.590 mol × 2 = 1.18 mol

Now, we can find the theoretical yield of ammonia in grams:

Theoretical Yield = (moles of NH₃) × (molar mass of NH₃) = 1.18 mol × 17.03 g/mol ≈ 20.06 g

To calculate the percent yield, we can use the formula:

Percent Yield = (Actual Yield / Theoretical Yield) × 100

Given that the actual yield is 8.33 grams, we can substitute these values into the formula:

Percent Yield = (8.33 g / 20.06 g) × 100 ≈ 41.5%

Therefore, the percent yield of ammonia is approximately 41.5%.

4: Heat Energy in Ammonia Production

Calculate moles of NH₃:

Moles = Mass / Molar Mass

Given that the molar mass of NH₃ is approximately 17.03 g/mol:

Moles = 8.33 g / 17.03 g/mol ≈ 0.489 mol

Use the coefficients in the balanced equation to relate moles of NH₃ to ΔH:

q = Moles × ΔHₙᵣₓₙ

q = 0.489 mol × ΔHₙᵣₓₙ

The sign of ΔHₙᵣₓₙ depends on whether the reaction is exothermic or endothermic.

5: Reaction of Ammonia with Nitric Acid

(a) Balanced Chemical Equation:

NH₃(g) + HNO₃(aq) → NH₄NO₃(aq)

(b) Complete Ionic Equation:

NH₃(g) + H⁺(aq) + NO₃⁻(aq) → NH₄⁺(aq) + NO₃⁻(aq)

Net Ionic Equation:

NH₃(g) + H⁺(aq) → NH₄⁺(aq)

In the net ionic equation, the nitrate ion (NO₃⁻) is a spectator ion and does not participate in the reaction. The net ionic equation focuses on the species that undergo a change.

User Laodao
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