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The question in the image asks: 

**"If 15.0 g of nitrogen (N₂) and 2.00 g of hydrogen (H₂) react to produce 1.38 g of ammonia (NH₃), according to the reaction below, what is the percent yield of the reaction?"**

The balanced chemical equation is:

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

### Steps to Solve:

1. **Determine the limiting reactant:**
   - Moles of nitrogen (N₂):
     \[
     \text{Molar mass of N₂} = 28.02 \, \text{g/mol}
     \]
     \[
     \text{Moles of N₂} = \frac{15.0 \, \text{g}}{28.02 \, \text{g/mol}} = 0.535 \, \text{mol}
     \]
   - Moles of hydrogen (H₂):
     \[
     \text{Molar mass of H₂} = 2.016 \, \text{g/mol}
     \]
     \[
     \text{Moles of H₂} = \frac{2.00 \, \text{g}}{2.016 \, \text{g/mol}} = 0.993 \, \text{mol}
     \]

2. **Determine which reactant is limiting:**
   According to the balanced equation, 1 mole of N₂ reacts with 3 moles of H₂. Therefore, for 0.535 mol of N₂, we need:
   \[
   0.535 \, \text{mol N₂} \times 3 = 1.605 \, \text{mol H₂}
   \]
   But we only have 0.993 mol of H₂, so H₂ is the limiting reactant.

3. **Calculate the theoretical yield of NH₃:**
   From the balanced equation, 3 moles of H₂ produce 2 moles of NH₃. Thus:
   \[
   0.993 \, \text{mol H₂} \times \frac{2 \, \text{mol NH₃}}{3 \, \text{mol H₂}} = 0.662 \, \text{mol NH₃}
   \]
   The molar mass of NH₃ is:
   \[
   \text{Molar mass of NH₃} = 17.03 \, \text{g/mol}
   \]
   Therefore, the theoretical yield is:
   \[
   0.662 \, \text{mol NH₃} \times 17.03 \, \text{g/mol} = 11.27 \, \text{g NH₃}
   \]

4. **Calculate the percent yield:**
   The actual yield is 1.38 g, so the percent yield is:
   \[
   \text{Percent yield} = \left( \frac{1.38 \, \text{g}}{11.27 \, \text{g}} \right) \times 100 = 12.2\%
   \]

### Answer:
The percent yield of the reaction is **12.2%**.

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Do you want further details on any step or more explanation? Here are five related questions for further understanding:

1. What is the significance of identifying the limiting reactant in a chemical reaction?
2. How would the percent yield change if the actual yield were higher?
3. Why is it important to have a balanced chemical equation for such calculations?
4. How does the concept of stoichiometry help in predicting the amounts of products formed?
5. What factors might cause the actual yield to be lower than the theoretical yield?

**Tip:** Always make sure to use the correct molar masses when calculating moles, as small errors can affect the entire calculation.

If 15.0 g of nitrogen (N₂) and 2.00 g of hydrogen (H₂) react to produce 1.38 g of ammonia (NH₃), according to the reaction below, what is the percent yield of the reaction?

Learn how to calculate the percent yield of ammonia in a chemical reaction where 15.0 g of nitrogen reacts with 2.00 g of hydrogen. This example involves stoichiometry, limiting reactants, and percent yield calculations.

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