Answer:
1) C9H16 + 13O2 → 9CO2 + 8H2O
2) C10H20 + 15O2 → 10CO2 + 10H2O
3) C8H20 + 13O2 → 8CO2 + 10H2O
Step-by-step explanation:
Complete combustion is when a substance burns and reacts with oxygen fully to produce carbon dioxide and water.
1) With the first molecule C9H16, let's write the equation:
C9H16 + O2 → CO2 + H2O
This equation is not balanced. First, let's balance the carbons in the reactants (left side) and products (right side):
We have 9 carbons on the left and 1 carbon on the right, so let's add a coefficient of 9 in front of the CO2 to make it balanced:
C9H16 + O2 → 9CO2 + H2O
Now that the number of carbon is balanced on both sides, let's balance the number of hydrogen:
The equation we have now:
C9H16 + O2 → 9CO2 + H2O
We have 16 hydrogens on the left and 2 hydrogens on the right, so let's add a coefficient of 8 in front of H20 because 8 × 2 = 16:
C9H16 + O2 → 9CO2 + 8H2O
All we need to do now is balance the oxygen:
We have 2 oxygens on the left and 26 oxygens on the right ( because in 9CO2 there are 9×2=18 oxygens and in 8H2O there are 8×1=8 oxygens, so 18+8=26)
Current equation:
C9H16 + O2 → 9CO2 + 8H2O (unbalanced)
To make the number of oxygens equal, add a coefficient of 13 in front of O2 because 13×2=26, so both sides are equal:
This is the final equation:
C9H16 + 13O2 → 9CO2 + 8H2O
2) Doing the same thing to C10H20:
C10H20 + O2 → CO2 + H2O
The number of carbons is not the same on both sides, so add a coefficient of 10 to CO2 so that there are 10 carbons on each side:
C10H20 + O2 → 10CO2 + H2O
Now let's balance the number of hydrogens. There are 20 hydrogens on the left and 2 hydrogens on the right, so add a coefficient of 10 in front of H2O because 10×2=20, so both sides have 20 hydrogens:
C10H20 + O2 → 10CO2 + 10H2O
Lastly balancing the number of oxygens. There are 2 oxygens on the left and 30 oxygens on the right (because in 10CO2 there are 10×2=20 oxygens and in 10H2O there are 10×1=10 oxygens, so 20+10=30)
Add a coefficient of 15 in front of O2 because 15×2=30, making both sides of oxygen equal to 30
Final equation:
C10H20 + 15O2 → 10CO2 + 10H2O
3) Now for C8H20:
C8H20 + O2 → CO2 + H2O (unbalanced)
Let's balance the number of carbons by adding a coefficient of 8 in front of CO2 to make both sides of carbon equal to 8:
C8H20 + O2 → 8CO2 + H2O
There are 20 hydrogens on the left and 2 hydrogens on the right, so add a coefficient of 10 to H2O because 10×2=20, so both sides of hydrogen are equal to 20:
C8H20 + O2 → 8CO2 + 10H2O
Lastly, balancing the oxygens. There are 2 oxygens on the left and 26 oxygens on the right (because in 8CO2 there are 8×2=16 oxygens and in 10H2O there are 10×1=10 oxygens, so 16+10=26)
Add a coefficient of 13 in front of O2 because 13×2=26, so there will be 26 oxygens on both sides.
Final equation:
C8H20 + 13O2 → 8CO2 + 10H2O