Nitrogen reacts with hydrogen to make ammonia.
a) Calculate the maximum theoretical mass of ammonia that can be made by reacting 90.0 g of hydrogen with an
excess of nitrogen.
b) In the reaction, only 153 g of ammonia was produced. Calculate the percentage yield.

Ask by Long Bowers. in Colombia

Jan 19,2025

Question

Nitrogen reacts with hydrogen to make ammonia.
a) Calculate the maximum theoretical mass of ammonia that can be made by reacting 90.0 g of hydrogen with an
excess of nitrogen.
b) In the reaction, only 153 g of ammonia was produced. Calculate the percentage yield.

Ask by Long Bowers. in Colombia

Jan 19,2025

UpStudy AI · Accepted Answer

Sure, let's solve both parts step-by-step.

### **Given Reaction:**
$$ \mathrm{N}_{2} + 3\, \mathrm{H}_{2} \rightarrow 2\, \mathrm{NH}_{3} $$

### **Part (a): Calculate the Maximum Theoretical Mass of Ammonia**

1. **Determine the Molar Masses:**
   - **Hydrogen ($$\mathrm{H}_2$$):**  
     $$ \mathrm{H}_2 = 2 \times 1.008\, \text{g/mol} = 2.016\, \text{g/mol} $$
   - **Ammonia ($$\mathrm{NH}_3$$):**  
     $$ \mathrm{NH}_3 = 14.007\, (\text{N}) + 3 \times 1.008\, (\text{H}) = 17.031\, \text{g/mol} $$

2. **Calculate Moles of Hydrogen:**
   $$ \text{Moles of } \mathrm{H}_2 = \frac{90.0\, \text{g}}{2.016\, \text{g/mol}} \approx 44.64\, \text{mol} $$

3. **Use Stoichiometry to Find Moles of $$\mathrm{NH}_3$$:**
   From the balanced equation, 3 moles of $$\mathrm{H}_2$$ produce 2 moles of $$\mathrm{NH}_3$$.
   $$ \text{Moles of } \mathrm{NH}_3 = \left(\frac{2}{3}\right) \times 44.64\, \text{mol} \approx 29.76\, \text{mol} $$

4. **Convert Moles of $$\mathrm{NH}_3$$ to Mass:**
   $$ \text{Mass of } \mathrm{NH}_3 = 29.76\, \text{mol} \times 17.031\, \text{g/mol} \approx 506.9\, \text{g} $$
   
   **Rounded to three significant figures:**
   $$ \text{Mass of } \mathrm{NH}_3 \approx 507\, \text{g} $$

### **Part (b): Calculate the Percentage Yield**

1. **Given:**
   - **Actual Yield of $$\mathrm{NH}_3$$:** 153 g

2. **Calculate Percentage Yield:**
   $$ \text{Percentage Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\% $$
   $$ \text{Percentage Yield} = \left(\frac{153\, \text{g}}{507\, \text{g}}\right) \times 100\% \approx 30.2\% $$

### **Final Answers:**

**a)** The maximum theoretical mass of ammonia produced is **507 g**.

**b)** The percentage yield is approximately **30.2 %**.
**a)** The maximum theoretical mass of ammonia produced is **507 g**. **b)** The percentage yield is approximately **30.2 %**.

Nitrogen reacts with hydrogen to make ammonia.
a) Calculate the maximum theoretical mass of ammonia that can be made by reacting 90.0 g of hydrogen with an
excess of nitrogen.
b) In the reaction, only 153 g of ammonia was produced. Calculate the percentage yield.

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Nitrogen reacts with hydrogen to make ammonia. a) Calculate the maximum theoretical mass of ammonia that can be made by reacting 90.0 g of hydrogen with an excess of nitrogen. b) In the reaction, only 153 g of ammonia was produced. Calculate the percentage yield.

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Latest Chemistry Questions

Nitrogen reacts with hydrogen to make ammonia.
a) Calculate the maximum theoretical mass of ammonia that can be made by reacting 90.0 g of hydrogen with an
excess of nitrogen.
b) In the reaction, only 153 g of ammonia was produced. Calculate the percentage yield.