Balance the following chemical equation, then answer the following question.

C8H18(g)+O2(g)→CO2(g)+H2O(g)

How many grams of oxygen are required to react with 12.0 grams of octane (C8H18) in the combustion of octane in gasoline?  

### Step 1: Balance the chemical equation

The unbalanced chemical equation is:

\[
\text{C}_8\text{H}_{18}(g) + \text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(g)
\]

To balance it:

1. **Carbon (C)**: There are 8 carbon atoms in C₈H₁₈, so we need 8 CO₂ molecules on the product side:
   \[
   \text{C}_8\text{H}_{18}(g) + \text{O}_2(g) \rightarrow 8\text{CO}_2(g) + \text{H}_2\text{O}(g)
   \]

2. **Hydrogen (H)**: There are 18 hydrogen atoms in C₈H₁₈, so we need 9 H₂O molecules (each H₂O has 2 H atoms):
   \[
   \text{C}_8\text{H}_{18}(g) + \text{O}_2(g) \rightarrow 8\text{CO}_2(g) + 9\text{H}_2\text{O}(g)
   \]

3. **Oxygen (O)**: On the right side, we now have 8 CO₂ molecules (16 O atoms) and 9 H₂O molecules (9 O atoms), giving a total of \(16 + 9 = 25\) oxygen atoms. Since each O₂ molecule provides 2 oxygen atoms, we need \( \frac{25}{2} = 12.5 \) O₂ molecules. 

4. To avoid fractional coefficients, multiply the entire equation by 2:
   \[
   2\text{C}_8\text{H}_{18}(g) + 25\text{O}_2(g) \rightarrow 16\text{CO}_2(g) + 18\text{H}_2\text{O}(g)
   \]

So the balanced equation is:

\[
2\text{C}_8\text{H}_{18}(g) + 25\text{O}_2(g) \rightarrow 16\text{CO}_2(g) + 18\text{H}_2\text{O}(g)
\]

### Step 2: Determine how many grams of oxygen (O₂) are required for 12.0 g of octane (C₈H₁₈)

1. **Molar mass of C₈H₁₈ (octane)**:
   \[
   \text{C}_8\text{H}_{18}: (8 \times 12.01 \, \text{g/mol}) + (18 \times 1.008 \, \text{g/mol}) = 114.23 \, \text{g/mol}
   \]

2. **Moles of octane**:
   \[
   \text{moles of C}_8\text{H}_{18} = \frac{12.0 \, \text{g}}{114.23 \, \text{g/mol}} = 0.105 \, \text{mol}
   \]

3. **Stoichiometric ratio from the balanced equation**: 
   According to the balanced equation, 2 moles of octane react with 25 moles of oxygen gas (O₂). So, the mole ratio is:
   \[
   \frac{25 \, \text{mol O}_2}{2 \, \text{mol C}_8\text{H}_{18}} = 12.5 \, \text{mol O}_2 \text{ per } 1 \, \text{mol C}_8\text{H}_{18}
   \]

4. **Moles of oxygen required**:
   \[
   0.105 \, \text{mol C}_8\text{H}_{18} \times 12.5 \, \text{mol O}_2 / \text{mol C}_8\text{H}_{18} = 1.31 \, \text{mol O}_2
   \]

5. **Molar mass of O₂**:
   \[
   \text{O}_2: (2 \times 16.00 \, \text{g/mol}) = 32.00 \, \text{g/mol}
   \]

6. **Mass of oxygen required**:
   \[
   1.31 \, \text{mol O}_2 \times 32.00 \, \text{g/mol} = 41.9 \, \text{g O}_2
   \]

### Final Answer:
41.9 grams of oxygen are required to react with 12.0 grams of octane in the combustion process.

---

Do you have any questions or need more details on any step?

Here are 5 related questions to expand your understanding:
1. What is the significance of balancing a chemical equation?
2. How do you determine the molar mass of a compound?
3. What is the mole-to-mole ratio, and why is it important in stoichiometry?
4. How would the amount of oxygen required change if you had 24 grams of octane?
5. What are the products of the combustion of hydrocarbons, and how do they impact the environment?

**Tip**: Always check if a chemical equation is balanced before proceeding with stoichiometric calculations. It ensures accurate conversion between moles of reactants and products.

Balance the following chemical equation, then answer the following question. C8H18(g) + O2(g) → CO2(g) + H2O(g). How many grams of oxygen are required to react with 12.0 grams of octane (C8H18) in the combustion of octane in gasoline?

Find out how to balance the combustion reaction of octane (C8H18) with oxygen (O2) and calculate the grams of oxygen required to react with 12.0 grams of octane. This stoichiometry problem includes detailed steps and explanations, ideal for high school chemistry students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation