10 of 10 Submit 1 Balancing Chemical Reactions Balance the reaction and answer the questions. $$ -\mathrm{C}_{2} \mathrm{H}_{6}+-\mathrm{O}_{2} \rightarrow-\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} $$ PART A The reaction here shows a simple chemical reaction involving the combustion of ethane $$ \left(\mathrm{C}_{2} \mathrm{H}_{6}\right) $$, a fuel, in the presence of oxygen gas to form water and carbon dioxide $$ \left(\mathrm{CO}_{2}\right) $$. Which coefficients can be placed in the blanks to correctly balance the equation? A $$ 1,3,2,3 $$ B $$ 2,7,4,6 $$ C $$ \quad 2,6,4,7 $$ D $$ \quad 2,4,3,2 $$ PART B

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To **balance the chemical reaction** for the combustion of ethane ($$\mathrm{C}_2\mathrm{H}_6$$), we'll determine the appropriate coefficients for each reactant and product to ensure that the number of atoms for each element is the same on both sides of the equation.

### **Given Unbalanced Equation:**
$$
\_\mathrm{C}_{2} \mathrm{H}_{6} + \_\mathrm{O}_{2} \rightarrow \_\mathrm{CO}_{2} + \_\mathrm{H}_{2} \mathrm{O}
$$

Let's assign coefficients $$a$$, $$b$$, $$c$$, and $$d$$ to each compound respectively:
$$
a\,\mathrm{C}_{2}\mathrm{H}_{6} + b\,\mathrm{O}_{2} \rightarrow c\,\mathrm{CO}_{2} + d\,\mathrm{H}_{2}\mathrm{O}
$$

### **Step-by-Step Balancing:**

1. **Carbon (C) Balance:**
   - **Left Side:** $$2a$$ atoms of C
   - **Right Side:** $$c$$ atoms of C
   $$
   2a = c \quad \text{(Equation 1)}
   $$

2. **Hydrogen (H) Balance:**
   - **Left Side:** $$6a$$ atoms of H
   - **Right Side:** $$2d$$ atoms of H
   $$
   6a = 2d \implies 3a = d \quad \text{(Equation 2)}
   $$

3. **Oxygen (O) Balance:**
   - **Left Side:** $$2b$$ atoms of O
   - **Right Side:** $$2c + d$$ atoms of O
   $$
   2b = 2c + d \quad \text{(Equation 3)}
   $$

4. **Substituting Equations 1 and 2 into Equation 3:**
   - From Equation 1: $$c = 2a$$
   - From Equation 2: $$d = 3a$$
   
   Substituting these into Equation 3:
   $$
   2b = 2(2a) + 3a \implies 2b = 4a + 3a \implies 2b = 7a \implies b = \frac{7a}{2}
   $$
   
5. **Choosing the Smallest Integer Value:**
   - To eliminate the fraction, let’s choose $$a = 2$$:
     $$
     b = \frac{7 \times 2}{2} = 7
     $$
   - Then:
     $$
     c = 2a = 4 \quad \text{and} \quad d = 3a = 6
     $$

### **Balanced Equation:**
$$
2\,\mathrm{C}_{2}\mathrm{H}_{6} + 7\,\mathrm{O}_{2} \rightarrow 4\,\mathrm{CO}_{2} + 6\,\mathrm{H}_{2}\mathrm{O}
$$

### **Verification:**
- **Carbon:** $$2 \times 2 = 4$$ on both sides.
- **Hydrogen:** $$2 \times 6 = 12$$ on both sides.
- **Oxygen:** $$7 \times 2 = 14$$ on the left; $$4 \times 2 + 6 \times 1 = 14$$ on the right.

### **Conclusion:**

**The correct set of coefficients is:**

**B) $$2, 7, 4, 6$$**
The correct coefficients to balance the equation are 2, 7, 4, and 6.

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