Balance each of the following skeleton equations. This is the last one left to balance.
D) C4H10 + O2 -> CO2 + H2O
= C4H10 + O2 -> 4CO2 + 10H
What you left is not balanced. Oxygen has 2 atoms on the left and 8 on the right. Second, H NEVER occurs as atomic H (unless we work on it to make it do it) but occurs as diatomic H as H2, Same for H2, O2, N2, Br2, Cl2, F2, I2 and their cousins P4 and S8.
2C4H10 + 13 O2 ==> 8CO2 + 10H2O
To balance the equation C4H10 + O2 -> CO2 + H2O, follow these steps:
1. Count the number of atoms of each element on both sides of the equation. In this case, we have:
- Carbon (C): 4 atoms on the left and 1 atom on the right
- Hydrogen (H): 10 atoms on the left and 2 atoms on the right
- Oxygen (O): 2 atoms on the left and 3 atoms on the right
2. Begin by balancing the elements that appear in the greatest number of compounds. In this case, it's carbon.
Since there are 4 carbon atoms on the left, we need to put a coefficient of 4 in front of CO2 on the right to balance the carbon atoms:
C4H10 + O2 -> 4CO2 + H2O
3. Next, balance the hydrogen atoms by adding a coefficient of 5 in front of H2O:
C4H10 + O2 -> 4CO2 + 5H2O
4. Finally, balance the oxygen atoms. We have 13 oxygen atoms on the right side of the equation, but only 2 on the left side.
Add a coefficient of 13/2 (which can be simplified as 6.5) in front of O2 to balance the oxygen atoms:
C4H10 + 6.5O2 -> 4CO2 + 5H2O
5. Balancing coefficients must be whole numbers (cannot be fractions). To eliminate the fraction, multiply through by 2 to make all the coefficients whole numbers:
2C4H10 + 13O2 -> 8CO2 + 10H2O
The balanced equation is:
2C4H10 + 13O2 -> 8CO2 + 10H2O