can anyone check and see if i balanced the equation for the combustion of benzene correctly, thank you.
C6H6 + O2 --> CO2 + H2O
Answer: C6H6 + 15O2 --> 6CO2 + 3H2O
It doesn't look right to me; however, you can check ALL balanced(?) equations and know when you are right or wrong.
You have 6C on the left and 6C on the right.
You have 6H on the left and 6 H on eht right.
You have 30 O on the left and 15 (6*2=12 from 6CO2 + 3 from 3H2O) on the right. So it isn't balanced. Generally, when you have everything balanced EXCEPT for one item sitting there by itself, the thing to do is to count up what you need on the right and MAKE it that number on the left. So, we need 15 O atoms on the right, we will need 7-1/2 O2 on the left. Since we usually don't balance with 1/2 of anything, just multiply everything through by 2 to get rid of the 1/2. That will do it I think.
2C6H6 + 15O2 -> 12CO2 + 6H20
I need answers
To balance the equation for the combustion of benzene, we start by checking the number of atoms on each side to see if they are equal.
On the left side of the equation, we have:
- 6 carbon (C6H6)
- 6 hydrogen (C6H6)
- 2 oxygen from the oxygen molecules (O2)
On the right side of the equation, we have:
- 6 carbon from the carbon dioxide (CO2)
- 3 hydrogen from the water (H2O)
- x oxygen from both carbon dioxide and water (CO2 + H2O)
To balance the carbons, hydrogens, and oxygens, we can see that they are already balanced. However, the equation is not balanced because we have 2 oxygen atoms on the left side and we need to have the same number on the right side.
To fix this, we want to make sure there is an equal number of oxygen atoms on both sides. Since we have 2 oxygen atoms on the left and we need 15 oxygen atoms (6 from CO2 + 3 from H2O) on the right, we will multiply everything by 15 to achieve this.
The balanced equation for the combustion of benzene is:
C6H6 + 15O2 --> 6CO2 + 3H2O
Now, there are 6 carbon atoms on both sides, 6 hydrogen atoms on both sides, and 15 oxygen atoms on both sides, making it a balanced equation.