Question

Calculate the number of moles of oxygen required to fully react with 500 grams of ethane (C2H6) in the combustion reaction 2C2H6 + 7O2 --> 4CO2 + 6H2O

Answer 1

To determine the number of moles of oxygen required to fully react with 500 grams of ethane (C₂H₆) in the combustion reaction \(2C₂H₆ + 7O₂ \rightarrow 4CO₂ + 6H₂O\), we can follow these steps:

1. **Calculate the molar mass of ethane (C₂H₆)**:
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.008 g/mol
- Molar mass of C₂H₆ = (2 × 12.01) + (6 × 1.008)
- Molar mass of C₂H₆ = 24.02 + 6.048 = 30.07 g/mol

2. **Calculate the number of moles of ethane in 500 grams**:
\[
\text{Moles of C₂H₆} = \frac{\text{mass}}{\text{molar mass}} = \frac{500 \text{ g}}{30.07 \text{ g/mol}} \approx 16.64 \text{ moles of C₂H₆}
\]

3. **Use the stoichiometry of the reaction to find moles of O₂ needed**:
The balanced reaction shows that 2 moles of C₂H₆ react with 7 moles of O₂.
Thus, the ratio of moles of O₂ to moles of C₂H₆ is:
\[
\frac{7 \text{ moles O₂}}{2 \text{ moles C₂H₆}}
\]

4. **Calculate moles of O₂ required for 16.64 moles of C₂H₆**:
\[
\text{Moles of O₂} = 16.64 \text{ moles C₂H₆} \times \frac{7 \text{ moles O₂}}{2 \text{ moles C₂H₆}}
\]
\[
\text{Moles of O₂} = 16.64 \times 3.5 = 58.24 \text{ moles O₂}
\]

Therefore, **the number of moles of oxygen required to fully react with 500 grams of ethane is approximately 58.24 moles of O₂**.