How do I balance chemical equations? Here's an example problem
C2H5OH +O2 --> CO2 + H20
Thanks!
I like to start with the most difficult.
Notice on the ethanol, 2 carebons, so put a 2 on the carbon dioxide.
C2H5OH >>2 CO2
next, the H in ethanol, there ares six. Go to the H2O and put a 3 on it.
C2H5OH >>2 CO2+ 3H2O
Now, the O . Notice one on the H2O, you have 3 (one 2 +1). So on the right, you have 4+2+1...which means you have to have six O from the O2 (one is coming from the OH).
C2H5OH + 3O2 >>2 CO2+ 3H2O
So what I am telling you, I start from the uglies compound, the work toward a solution.
C2H5OH >>2 CO2+ 3H2O
Balance the reactions 1)C2H5OH+O2=Co2+H2O 2)C4H10+O2=Co2+H2O
To balance a chemical equation, you need to ensure that the same number of atoms of each element appear on both sides of the equation. Here's a step-by-step guide on how to balance the given chemical equation:
Step 1: Count the number of atoms for each element on both sides of the equation.
On the left side:
C: 2 atoms
H: 6 atoms
O: 1 atom
On the right side:
C: 1 atom
H: 2 atoms
O: 3 atoms
Step 2: Begin by balancing the most complex or abundant atoms. Let's start with carbon (C) in this case.
Step 3: Add coefficients (whole numbers) in front of the molecules to balance the number of atoms on both sides. In this case, add a coefficient of 2 in front of CO2 on the right side of the equation.
C2H5OH + O2 --> 2CO2 + H2O
Now, review the counts of each element:
On the left side:
C: 2 atoms
H: 6 atoms
O: 1 atom
On the right side:
C: 2 atoms
H: 2 atoms
O: 5 atoms
Step 4: Next, balance the hydrogen (H) atoms. There are 6 hydrogen atoms on the left side and only 2 on the right side. To balance this, add a coefficient of 3 in front of H2O on the right side.
C2H5OH + O2 --> 2CO2 + 3H2O
Now, review the counts of each element once again:
On the left side:
C: 2 atoms
H: 6 atoms
O: 1 atom
On the right side:
C: 2 atoms
H: 6 atoms
O: 9 atoms
Step 5: Finally, balance the oxygen (O) atoms. There are 1 oxygen atom on the left side and 9 oxygen atoms on the right side. To balance this, add a coefficient of 5/2 or 2.5 in front of O2 on the left side. However, we cannot use fractional coefficients in balanced chemical equations. Multiply all coefficients by 2 to remove the fraction:
2C2H5OH + 5O2 --> 4CO2 + 6H2O
Now, all elements are balanced:
On the left side:
C: 4 atoms
H: 10 atoms
O: 10 atoms
On the right side:
C: 4 atoms
H: 12 atoms
O: 12 atoms
By following these steps, the given chemical equation C2H5OH + O2 -> CO2 + H2O has been balanced to 2C2H5OH + 5O2 -> 4CO2 + 6H2O.
The number of each element on the reactant side must equal the number of elements on the product side. Such gives rise the the law of mass balance which states ....
Sum of Mass of Reactants = Sum of Mass of Products
Now, for Oxidation-Reduction Reactions you have to have both a charge transfer balance and a mass balance. That is, the no of electrons used in the oxidation process must equal the number of electrons in the reduction process.
Balancing a chemical equation is sometimes quite simple and can be done by inspection while with some equations rules of order must be applied. The example submitted can easily be balanced by simple inspection provided you conform to some basic practical rules. That is...
=> Only 'coefficient' numbers can be used to balance elements within an equation. Subscripts can not be changed as such would change the identity of the substance in the reactions,
=> For reactions containing multiple polyions, balance the polyions if applicable, not each element in the polyion. Example...
........ Pb(NO₃)₂ + 2NaCl => 2NaNO₃ + PbCl₂
2 NO₃ˉ on reactant side & 2 NO₃ˉ on product side
=> Save elements 'by themselves' until last to balance. Fractions may be used to complete the balance, then multiply by the denominator of the fraction used to convert coefficients to whole numbers.
=> Also, by convention, one should reduce the coefficients of the balanced equation to the smallest whole number values. This is what is referred to as 'The Standard Equation' and is assumed to be under conditions of 0-deg Celcius and 1 atm pressure.
For the posted equation ...
Start with the carbons, then the hydrogens and finally the oxygens.
C2H5OH + O2 --> 2CO2 + H20 Increase CO2 to 2CO2 => 2C = 2C
C2H5OH + O2 --> 2CO2 + 3H20 Increase H2O to 3H2O => 6H = 6 H
C2H5OH + 3O2 --> 2CO2 + 3H20 Increase O2 to 3O2 => 7O = 7O