Use dimensional analysis to show all work: How many grams of excess reagent is used and how many grams remain?  A student made 102 g of carbon dioxide what is her % yield?  (Rest of work provided on images.)

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Use dimensional analysis to show all work:

How many grams of excess reagent is used and how many grams remain? 

A student made 102 g of carbon dioxide what is her % yield? 

(Rest of work provided on images.)

### Combustion Reaction of C₆H₁₀

1. **Chemical Equation:**
   \[
   \text{C}_6\text{H}_{10(g)} + \frac{17}{2} \text{O}_{2(g)} \xrightarrow{\Delta} 6\text{CO}_{2(g)} + 5\text{H}_2\text{O}_{(g)}
   \]

### Calculations

2. **For C₆H₁₀:**
   - Molar mass of C₆H₁₀: 
     - \(1 \, \text{mol} = 82 \, \text{g}\)
     - Conversion factor: \(\frac{1 \, \text{mol}}{82 \, \text{g}} = 1\)
     - \(\frac{82 \, \text{g}}{1 \, \text{mol}} = 1\)

   - Converting grams to moles:
     \[
     45 \, \text{g} \times \frac{1 \, \text{mol}}{82 \, \text{g}} = 0.548 \, \text{mol}
     \]

3. **For O₂:**
   - Molar mass of O₂:
     - \(1 \, \text{mol} = 32 \, \text{g}\)
     - Conversion factor: \(\frac{1 \, \text{mol}}{32 \, \text{g}} = 1\)
     - \(\frac{32 \, \text{g}}{1 \, \text{mol}} = 1\)

   - Converting grams to moles:
     \[
     156 \, \text{g} \times \frac{1 \, \text{mol}}{32 \, \text{g}} = 4.875 \, \text{mol}
     \]
Transcribed Image Text:### Combustion Reaction of C₆H₁₀ 1. **Chemical Equation:** \[ \text{C}_6\text{H}_{10(g)} + \frac{17}{2} \text{O}_{2(g)} \xrightarrow{\Delta} 6\text{CO}_{2(g)} + 5\text{H}_2\text{O}_{(g)} \] ### Calculations 2. **For C₆H₁₀:** - Molar mass of C₆H₁₀: - \(1 \, \text{mol} = 82 \, \text{g}\) - Conversion factor: \(\frac{1 \, \text{mol}}{82 \, \text{g}} = 1\) - \(\frac{82 \, \text{g}}{1 \, \text{mol}} = 1\) - Converting grams to moles: \[ 45 \, \text{g} \times \frac{1 \, \text{mol}}{82 \, \text{g}} = 0.548 \, \text{mol} \] 3. **For O₂:** - Molar mass of O₂: - \(1 \, \text{mol} = 32 \, \text{g}\) - Conversion factor: \(\frac{1 \, \text{mol}}{32 \, \text{g}} = 1\) - \(\frac{32 \, \text{g}}{1 \, \text{mol}} = 1\) - Converting grams to moles: \[ 156 \, \text{g} \times \frac{1 \, \text{mol}}{32 \, \text{g}} = 4.875 \, \text{mol} \]
### Reaction Calculations: Limiting Reactant Analysis

In this calculation, we’re analyzing the reaction involving C₆H₁₀ and O₂.

#### Step-by-Step Analysis

1. **Initial Reaction Setup**
   - From the reaction: 
     - 1 mol C₆H₁₀ reacts with \(\frac{17}{2}\) mol O₂.
   - Therefore, 0.548 mol C₆H₁₀ reacts with:
     \[
     \frac{17}{2} \times 0.548 \, \text{mol O}_2 = 4.658 \, \text{mol O}_2
     \]

2. **Determine the Limiting Reactant**
   - The limiting reactant is identified as C₆H₁₀.

3. **Product Calculation**
   - 1 mol C₆H₁₀ produces 5 mol CO₂.
   - Therefore, 0.548 mol C₆H₁₀ produces:
     \[
     5 \times 0.548 \, \text{mol CO}_2 = 2.74 \, \text{mol CO}_2
     \]

4. **Convert Moles of CO₂ to Grams**
   - The molar mass of CO₂ is 44 g/mol.
   - Converting 2.74 mol CO₂ to grams:
     \[
     2.74 \, \text{mol} \times \frac{44 \, \text{g}}{1 \, \text{mol}} = 120.56 \, \text{g}
     \]

5. **Conclusion**
   - (c) The limiting reactant is C₆H₁₀.

This calculation demonstrates how to identify the limiting reactant and determine the amount of product formed in a chemical reaction.
Transcribed Image Text:### Reaction Calculations: Limiting Reactant Analysis In this calculation, we’re analyzing the reaction involving C₆H₁₀ and O₂. #### Step-by-Step Analysis 1. **Initial Reaction Setup** - From the reaction: - 1 mol C₆H₁₀ reacts with \(\frac{17}{2}\) mol O₂. - Therefore, 0.548 mol C₆H₁₀ reacts with: \[ \frac{17}{2} \times 0.548 \, \text{mol O}_2 = 4.658 \, \text{mol O}_2 \] 2. **Determine the Limiting Reactant** - The limiting reactant is identified as C₆H₁₀. 3. **Product Calculation** - 1 mol C₆H₁₀ produces 5 mol CO₂. - Therefore, 0.548 mol C₆H₁₀ produces: \[ 5 \times 0.548 \, \text{mol CO}_2 = 2.74 \, \text{mol CO}_2 \] 4. **Convert Moles of CO₂ to Grams** - The molar mass of CO₂ is 44 g/mol. - Converting 2.74 mol CO₂ to grams: \[ 2.74 \, \text{mol} \times \frac{44 \, \text{g}}{1 \, \text{mol}} = 120.56 \, \text{g} \] 5. **Conclusion** - (c) The limiting reactant is C₆H₁₀. This calculation demonstrates how to identify the limiting reactant and determine the amount of product formed in a chemical reaction.
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