A compound containing 3 atom of carbon and 8 atoms of hydrogen is combined in a reaction wih oxygen molecules. The two end products of this equation are carbon dioxide (CO2)and water. What element should you look at first in balancing this equation?
Follow the instructions of your teacher; however, in real practice, it doesn't matter if you choose H or C first. I look at oxygen last.
To balance the equation, you should start by looking at the element that appears only once on each side of the equation. In this case, that element is carbon.
The first step is to count the number of carbon atoms on each side of the equation. On the reactant side, there are 3 carbon atoms, while on the product side, there is only 1 carbon atom in carbon dioxide (CO2).
To balance the carbon, you need to make the number of carbon atoms the same on both sides of the equation. You can achieve this by placing a coefficient of 3 in front of CO2, which will give you:
C3H8 + __O2 -> 3CO2 + H2O
Now, the carbon is balanced with 3 carbon atoms on both sides of the equation.
The next step is to balance the hydrogen atoms. On the reactant side, there are 8 hydrogen atoms, while on the product side, there are only 2 hydrogen atoms in water (H2O).
To balance the hydrogen, you need to place a coefficient of 4 in front of H2O, which will give you:
C3H8 + __O2 -> 3CO2 + 4H2O
Now, the number of hydrogen atoms is balanced with 8 hydrogen atoms on both sides of the equation.
Finally, you can balance the remaining oxygen atoms. On the reactant side, we have 2 oxygen atoms in each oxygen molecule (O2), so the total number of oxygen atoms is 2 times the coefficient for O2. On the product side, we have 6 oxygen atoms in 3CO2 and 4 oxygen atoms in 4H2O, giving a total of 10 oxygen atoms.
To balance the oxygen, you need to place a coefficient of 5 in front of O2, which will give you:
C3H8 + 5O2 -> 3CO2 + 4H2O
Now, the equation is balanced with 3 carbon atoms, 8 hydrogen atoms, and 10 oxygen atoms on both sides of the equation.