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------------------------------------------------ What is the maximum number of grams of ammonia, NH₃, which can be obtained from the reaction of 10.0 g of H₂ and 80.0 g of N₂?

\[ \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \]

A. 48.6 g
B. 97.3 g
C. 90.0 g
D. 56.3 g
E. 34.1 g

Answer :

The correct answer is option D. 56.3 g

To solve this problem, we need to determine the limiting reactant and then calculate the maximum amount of ammonia (NH₃) that can be produced based on the given quantities of H₂ and N₂.

Given information:

- Reaction equation: N₂ + 3H₂ → 2NH₃

- Mass of H₂ = 10.0 g

- Mass of N₂ = 80.0 g

Step 1: Convert the given masses to moles.

Molar mass of H₂ = 2.016 g/mol

Moles of H₂ = 10.0 g / 2.016 g/mol = 4.96 mol

Molar mass of N₂ = 28.014 g/mol

Moles of N₂ = 80.0 g / 28.014 g/mol = 2.85 mol

Step 2: Determine the limiting reactant.

According to the balanced equation, 3 moles of H₂ react with 1 mole of N₂.

Moles of H₂ required for 2.85 mol of N₂ = 2.85 mol × 3 = 8.55 mol

Moles of N₂ required for 4.96 mol of H₂ = 4.96 mol / 3 = 1.65 mol

Since there are only 4.96 mol of H₂ available, but 8.55 mol is required, H₂ is the limiting reactant.

Step 3: Calculate the maximum amount of NH₃ that can be produced.

According to the balanced equation, 2 moles of NH₃ are produced for every 3 moles of H₂ consumed.

Moles of NH₃ produced = (4.96 mol H₂ / 3) × 2 = 3.31 mol

Step 4: Convert moles of NH₃ to grams.

Molar mass of NH₃ = 17.031 g/mol

Mass of NH₃ = 3.31 mol × 17.031 g/mol = 56.3 g

Therefore, the maximum amount of ammonia (NH₃) that can be obtained from the reaction of 10.0 g of H₂ and 80.0 g of N₂ is 56.3 g.

The correct answer is option D. 56.3 g