In winemaking, the sugars in grapes undergo fermentation by yeast to yield CH3CH2OH and CO2. During cellular respiration, sugar and enthanol are "burned" to water vapor andCO2.
Write a combustion reaction for ethanol.
I need help with balancing the equation
i think it's
C2H6OH(l) + 3O2(g) ---> 2CO2(g) + 3H2O(g)
am I correct?..thanks!•chemistry - paul, Sunday, October 25, 2015 at 3:06pm
looks good to me, except the C2H6OH should be C2H5OH.
comment:
I did that, but it said...try again!
then it said HINT:
Start by writing a balanced chemical equation for the fermentation reaction of C6H12O6 and for the respiration reaction (combustion) of C6H12O6...what formula should be used to find the standard enthalpy of a reaction? Use the standard enthalpy of formation values from your textbook when calculating the standard enthalpy of reaction for both fermentation and respiration. thanks!
To write the balanced combustion reaction for ethanol, you need to consider the number of atoms on both sides of the equation. Here's how you can balance it correctly:
1. Start with the unbalanced equation: C2H5OH + O2 -> CO2 + H2O
2. Count the atoms on each side of the equation:
On the left side: 2 C, 6 H, and 1 O
On the right side: 1 C, 2 O, and 2 H
3. Balance the carbon atoms by applying a coefficient of 2 to the CO2:
C2H5OH + O2 -> 2CO2 + H2O
4. Now, let's balance the hydrogen atoms:
On the left side: 6 H
On the right side: 2 H
We need to balance the hydrogen atoms by adjusting the water molecules. To achieve this, we add the coefficient 3 in front of H2O:
C2H5OH + O2 -> 2CO2 + 3H2O
5. Finally, balance the oxygen atoms:
On the left side: 2 O
On the right side: 8 O (2 from CO2 and 6 from H2O)
To balance the oxygen, adjust the coefficient in front of O2 by multiplying it by 8/2 = 4:
C2H5OH + 4O2 -> 2CO2 + 3H2O
So, the balanced chemical equation for the combustion of ethanol is:
C2H5OH + 4O2 -> 2CO2 + 3H2O
Remember, when balancing chemical equations, it's important to ensure that the number of atoms is the same on both sides of the equation.