Balanced equation of the reaction that occurs when the butane in cigarette lighters burns in the presence of enough oxygen
2C4H10 + 13O2 -> 10H2O +8CO2
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To determine the balanced equation for the combustion of butane (C4H10) in the presence of oxygen (O2), we need to consider the balanced ratio of carbon, hydrogen, and oxygen atoms in the reactants and products.
The combustion of butane involves the reaction of butane (C4H10) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O) as the main products.
Step 1: Write the unbalanced equation:
C4H10 + O2 → CO2 + H2O
Step 2: Balance the carbon (C) atoms:
There are four carbon atoms on the left side and one carbon atom on the right side. To balance the carbon, we need to place a coefficient of 4 in front of the carbon dioxide (CO2) product.
C4H10 + O2 → 4CO2 + H2O
Step 3: Balance the hydrogen (H) atoms:
There are ten hydrogen atoms on the left side and two hydrogen atoms on the right side. To balance hydrogen, place a coefficient of 5 in front of the water (H2O) product.
C4H10 + O2 → 4CO2 + 5H2O
Step 4: Balance the oxygen (O) atoms:
There are two oxygen atoms in butane (C4H10) and two oxygen atoms in water (H2O) which total to four oxygen atoms. On the right side, there are four oxygen atoms in carbon dioxide (CO2). Hence, the oxygen atoms are already balanced.
C4H10 + O2 → 4CO2 + 5H2O
The balanced equation for the combustion of butane in the presence of enough oxygen is:
C4H10 + 13/2 O2 → 4CO2 + 5H2O
Note that the coefficient "13/2" is used for O2 to balance the oxygen atoms. It is common practice to balance fractional coefficients by multiplying the entire equation by a factor to convert them into whole numbers.