To determine the mass of $NH_3$ produced, we first need to use the stoichiometry of the reaction:

The balanced chemical equation is:

$N_2 + 3H_2 \rightarrow 2NH_3$

Step 1: Calculate the moles of $N_2$ and $H_2$

• Molar mass of $N_2 = 28 \, \text{g/mol}$ $\text{Moles of } N_2 = \frac{\text{Mass of } N_2}{\text{Molar mass of } N_2} = \frac{33.6}{28} = 1.2 \, \text{mol}$

• Molar mass of $H_2 = 2 \, \text{g/mol}$ $\text{Moles of } H_2 = \frac{\text{Mass of } H_2}{\text{Molar mass of } H_2} = \frac{1.2}{2} = 0.6 \, \text{mol}$

Step 2: Determine the limiting reagent

From the balanced equation, 1 mole of $N_2$ reacts with 3 moles of $H_2$. Therefore: $\text{Moles of } H_2 \text{ required for 1.2 moles of } N_2 = 1.2 \times 3 = 3.6 \, \text{mol}$

Since only 0.6 moles of $H_2$ are available, $H_2$ is the limiting reagent.

Step 3: Calculate the moles of $NH_3$ produced

From the balanced equation, 3 moles of $H_2$ produce 2 moles of $NH_3$. Therefore: $\text{Moles of } NH_3 = \frac{2}{3} \times \text{Moles of } H_2 = \frac{2}{3} \times 0.6 = 0.4 \, \text{mol}$

Step 4: Calculate the mass of $NH_3$

Molar mass of $NH_3 = 17 \, \text{g/mol}$: $\text{Mass of } NH_3 = \text{Moles of } NH_3 \times \text{Molar mass of } NH_3 = 0.4 \times 17 = 6.8 \, \text{g}$

Final Answer:

The mass of $NH_3$ produced is 6.8 g.

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Do you want details or have any questions?
Here are five related questions to expand your understanding:

1. How is the concept of limiting reagents applied in other chemical reactions?
2. What happens to the excess reagent in this reaction?
3. How can you calculate the theoretical yield of a chemical reaction?
4. How is stoichiometry useful in industrial chemical processes?
5. How does the balanced equation relate to mole ratios?

Tip: Always identify the limiting reagent