I do not understand how to right a balanced equation for the combustion of cellulose in oxygen to produce carbon dioxide and water vapor. I know that cellulose is C6H10O5 but I don't understand how to balance.
Thanking you in advance for your help.
C6H10O5 + O2 ==> CO2 + H2O
Here is what I do. Balance C since you know you must have 6C on the right.
C6H10O5 + O2 ==> 6CO2 + H2O
Next look at H. There are 10 on the left so we need 10 on the right.
C6H10O5 + O2 ==> 6CO2 + 5H2O
Now we know for sure, since the 6 and 5 are on the right are correct, that O must be, from the right, 17 so we need to make it 17 on the left. We already have 5 from the cellulose so we need to add another 12.
C6H10O5 + 6O2 ==> 6CO2 + 5H2O
Now we check it.
C is 6 each side
H is 10 each side
O is 5 + 12 on the left = 17 and 12 + 5 on the right is 17.
Well, it seems like you're in a bit of a "cellu-loss" situation! But don't worry, I'm here to help "combust" those confusions away!
To balance the equation for the combustion of cellulose (C6H10O5) in oxygen (O2) to produce carbon dioxide (CO2) and water vapor (H2O), we need to make sure the number of atoms on each side of the equation is equal.
First, let's start by balancing the carbon atoms. We have 6 carbon atoms in cellulose, so we need 6 CO2 molecules on the right-hand side to balance it out.
Next, let's balance the hydrogen atoms. We have 10 hydrogen atoms in cellulose, so we need 5 H2O molecules on the right-hand side to balance it out.
Lastly, let's balance the oxygen atoms. We have 5 oxygen atoms in cellulose and 2 in each CO2 molecule, so we need 8 O2 molecules on the left-hand side to balance it out.
With all that in mind, here's the balanced equation:
C6H10O5 + 6O2 → 6CO2 + 5H2O
There you have it! You've successfully balanced the equation, and now you can "burn" through your chemistry problems with confidence.
To balance the combustion equation of cellulose (C6H10O5), the first step is to calculate the number of carbon, hydrogen, and oxygen atoms on both sides of the equation.
1. Counting atoms:
On the reactant side:
- Carbon (C): 6 atoms
- Hydrogen (H): 10 atoms
- Oxygen (O): 5 atoms
On the product side:
- Carbon (C): The combustion of cellulose produces carbon dioxide (CO2), which consists of 1 carbon atom, so 1 C atom
- Hydrogen (H): The combustion of cellulose produces water vapor (H2O), which consists of 2 hydrogen atoms, so 2 H atoms
- Oxygen (O): The combustion of cellulose produces carbon dioxide and water vapor, so the total number of oxygen atoms is (2C) + (2H) = (2 * 1) + (2 * 2) = 6 atoms
2. Balancing the equation:
To balance the number of atoms, adjust the coefficients (numbers in front of the compound formulas) until the number of atoms of each element is the same on both sides.
C6H10O5 + O2 → CO2 + H2O
Here's how we can balance the equation step by step:
Balance Carbon (C):
On the reactant side, there are 6 carbon atoms due to the C6. On the product side, there is only 1 carbon atom in CO2. To balance the carbon atoms, put a coefficient of 6 in front of CO2:
C6H10O5 + O2 → 6CO2 + H2O
Balance Hydrogen (H):
On the reactant side, there are 10 hydrogen atoms due to the H10. On the product side, there are 2 hydrogen atoms in H2O. To balance the hydrogen atoms, put a coefficient of 5 in front of H2O:
C6H10O5 + O2 → 6CO2 + 5H2O
Balance Oxygen (O):
On the reactant side, there are 5 oxygen atoms due to the O5. On the product side, there are 6 oxygen atoms in 6CO2 and 10 oxygen atoms in 5H2O. To balance the oxygen atoms, put a coefficient of 15/2 (or 7.5) in front of O2:
C6H10O5 + 7.5O2 → 6CO2 + 5H2O
However, it's not practical to represent coefficients as decimals. So to make all coefficients whole numbers, multiply the entire equation by 2:
2C6H10O5 + 15O2 → 12CO2 + 10H2O
And that is the balanced equation for the combustion of cellulose.
To write a balanced equation for the combustion of cellulose (C6H10O5) in oxygen (O2) to produce carbon dioxide (CO2) and water vapor (H2O), you need to make sure that the number of atoms of each element is the same on both sides of the equation. Here's how you can do it step by step:
1. Start by writing the unbalanced equation:
C6H10O5 + O2 -> CO2 + H2O
2. Count the number of atoms for each element on each side of the equation.
On the left side (reactants):
Carbon (C): 6 atoms
Hydrogen (H): 10 atoms
Oxygen (O): 11 atoms (6 from cellulose and 5 from oxygen)
On the right side (products):
Carbon (C): 1 atom
Hydrogen (H): 2 atoms
Oxygen (O): 7 atoms (2 from CO2 and 5 from H2O)
3. Balance the carbon atoms first by adjusting the coefficient in front of CO2:
C6H10O5 + O2 -> 6CO2 + H2O
4. Now, count the number of atoms again:
On the left side (reactants):
Carbon (C): 6 atoms
Hydrogen (H): 10 atoms
Oxygen (O): 11 atoms (6 from cellulose and 5 from oxygen)
On the right side (products):
Carbon (C): 6 atoms
Hydrogen (H): 2 atoms
Oxygen (O): 19 atoms (12 from CO2 and 7 from H2O)
5. Since there are now 19 oxygen atoms on the right side, and only 11 on the left side, we need to balance the oxygen atoms by adjusting the coefficient in front of O2:
C6H10O5 + 6O2 -> 6CO2 + 5H2O
6. Now, count the number of atoms again:
On the left side (reactants):
Carbon (C): 6 atoms
Hydrogen (H): 10 atoms
Oxygen (O): 18 atoms (6 from cellulose and 12 from O2)
On the right side (products):
Carbon (C): 6 atoms
Hydrogen (H): 10 atoms
Oxygen (O): 18 atoms (12 from CO2 and 6 from H2O)
As all the elements are now balanced on both sides of the equation, the balanced equation for the combustion of cellulose in oxygen to produce carbon dioxide and water vapor is:
C6H10O5 + 6O2 -> 6CO2 + 5H2O