Answer: <u>2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g)</u>
Explanation:
Apologies for the error. Let me provide the correct balanced equation for the combustion of C2H6 (ethane) and address your additional question:
1. Combustion of Ethane (C2H6):
C2H6(g) + 3.5 O2(g) → 2 CO2(g) + 3 H2O(g)
To remove the half-coefficient, you can multiply the entire equation by 2:
<u>2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g)</u>
2. Combustion of Propane (C3H8):
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)
3. Combustion of Butane (C4H10):
C4H10(g) + 6.5 O2(g) → 4 CO2(g) + 5 H2O(g)
In all these reactions, the hydrocarbon (ethane, propane, butane) reacts with oxygen gas (O2) to produce carbon dioxide (CO2) and water (H2O).2C2H6(g)+7O2(g)→4CO2(g)+6H2O(g)