Answer :
To balance the combustion reaction of ethanol [tex]\((\mathrm{C_2H_5OH})\)[/tex], follow these steps:
1. Write down the unbalanced equation:
[tex]\[ \mathrm{C_2H_5OH} + \mathrm{O_2} \longrightarrow \mathrm{CO_2} + \mathrm{H_2O} \][/tex]
2. Count the number of atoms for each element on both sides of the reaction:
- Carbon (C):
- Left: [tex]\(2\)[/tex] (from ethanol, [tex]\(\mathrm{C_2H_5OH}\)[/tex])
- Right: [tex]\(1\)[/tex] (in [tex]\(\mathrm{CO_2}\)[/tex])
- Hydrogen (H):
- Left: [tex]\(6\)[/tex] (from ethanol, [tex]\(\mathrm{C_2H_5OH}\)[/tex])
- Right: [tex]\(2\)[/tex] (in [tex]\(\mathrm{H_2O}\)[/tex])
- Oxygen (O):
- Left: [tex]\(1\)[/tex] (from ethanol) + [tex]\(2\)[/tex] (from [tex]\(\mathrm{O_2}\)[/tex])
- Right: [tex]\(2\)[/tex] (in [tex]\(\mathrm{CO_2}\)[/tex]) + [tex]\(1\)[/tex] (in [tex]\(\mathrm{H_2O}\)[/tex])
3. Balance the number of Carbon atoms:
Since there are 2 carbon atoms in ethanol and only 1 in carbon dioxide, place a coefficient of 2 in front of [tex]\(\mathrm{CO_2}\)[/tex]:
[tex]\[ \mathrm{C_2H_5OH} + \mathrm{O_2} \longrightarrow 2 \mathrm{CO_2} + \mathrm{H_2O} \][/tex]
4. Balance the number of Hydrogen atoms:
There are 6 hydrogen atoms in ethanol and only 2 in water, thus placing a coefficient of 3 in front of [tex]\(\mathrm{H_2O}\)[/tex]:
[tex]\[ \mathrm{C_2H_5OH} + \mathrm{O_2} \longrightarrow 2 \mathrm{CO_2} + 3 \mathrm{H_2O} \][/tex]
5. Balance the number of Oxygen atoms:
Count the oxygen atoms on each side to balance them:
- Left side: [tex]\(1\)[/tex] (from ethanol) + [tex]\(2b\)[/tex] (oxygen molecules)
- Right side: [tex]\(2 \times 2 = 4\)[/tex] (from [tex]\(\mathrm{CO_2}\)[/tex]) + [tex]\(3\)[/tex] (from [tex]\(\mathrm{H_2O}\)[/tex])
Setting the total oxygen atoms equal on both sides:
[tex]\( 1 + 2b = 4 + 3 \)[/tex]
Simplifies to:
[tex]\( 1 + 2b = 7 \)[/tex]
[tex]\( 2b = 6 \)[/tex]
[tex]\( b = 3 \)[/tex]
Therefore, the equation is:
[tex]\[ \mathrm{C_2H_5OH} + 3\mathrm{O_2} \longrightarrow 2\mathrm{CO_2} + 3\mathrm{H_2O} \][/tex]
6. Conclusion:
The balanced combustion equation for ethanol is:
[tex]\[ \mathrm{C_2H_5OH} + 3\mathrm{O_2} \longrightarrow 2\mathrm{CO_2} + 3\mathrm{H_2O} \][/tex]
So, the final coefficients are:
[tex]\[ a = 1, \quad b = 3, \quad c = 2, \quad d = 3 \][/tex]
1. Write down the unbalanced equation:
[tex]\[ \mathrm{C_2H_5OH} + \mathrm{O_2} \longrightarrow \mathrm{CO_2} + \mathrm{H_2O} \][/tex]
2. Count the number of atoms for each element on both sides of the reaction:
- Carbon (C):
- Left: [tex]\(2\)[/tex] (from ethanol, [tex]\(\mathrm{C_2H_5OH}\)[/tex])
- Right: [tex]\(1\)[/tex] (in [tex]\(\mathrm{CO_2}\)[/tex])
- Hydrogen (H):
- Left: [tex]\(6\)[/tex] (from ethanol, [tex]\(\mathrm{C_2H_5OH}\)[/tex])
- Right: [tex]\(2\)[/tex] (in [tex]\(\mathrm{H_2O}\)[/tex])
- Oxygen (O):
- Left: [tex]\(1\)[/tex] (from ethanol) + [tex]\(2\)[/tex] (from [tex]\(\mathrm{O_2}\)[/tex])
- Right: [tex]\(2\)[/tex] (in [tex]\(\mathrm{CO_2}\)[/tex]) + [tex]\(1\)[/tex] (in [tex]\(\mathrm{H_2O}\)[/tex])
3. Balance the number of Carbon atoms:
Since there are 2 carbon atoms in ethanol and only 1 in carbon dioxide, place a coefficient of 2 in front of [tex]\(\mathrm{CO_2}\)[/tex]:
[tex]\[ \mathrm{C_2H_5OH} + \mathrm{O_2} \longrightarrow 2 \mathrm{CO_2} + \mathrm{H_2O} \][/tex]
4. Balance the number of Hydrogen atoms:
There are 6 hydrogen atoms in ethanol and only 2 in water, thus placing a coefficient of 3 in front of [tex]\(\mathrm{H_2O}\)[/tex]:
[tex]\[ \mathrm{C_2H_5OH} + \mathrm{O_2} \longrightarrow 2 \mathrm{CO_2} + 3 \mathrm{H_2O} \][/tex]
5. Balance the number of Oxygen atoms:
Count the oxygen atoms on each side to balance them:
- Left side: [tex]\(1\)[/tex] (from ethanol) + [tex]\(2b\)[/tex] (oxygen molecules)
- Right side: [tex]\(2 \times 2 = 4\)[/tex] (from [tex]\(\mathrm{CO_2}\)[/tex]) + [tex]\(3\)[/tex] (from [tex]\(\mathrm{H_2O}\)[/tex])
Setting the total oxygen atoms equal on both sides:
[tex]\( 1 + 2b = 4 + 3 \)[/tex]
Simplifies to:
[tex]\( 1 + 2b = 7 \)[/tex]
[tex]\( 2b = 6 \)[/tex]
[tex]\( b = 3 \)[/tex]
Therefore, the equation is:
[tex]\[ \mathrm{C_2H_5OH} + 3\mathrm{O_2} \longrightarrow 2\mathrm{CO_2} + 3\mathrm{H_2O} \][/tex]
6. Conclusion:
The balanced combustion equation for ethanol is:
[tex]\[ \mathrm{C_2H_5OH} + 3\mathrm{O_2} \longrightarrow 2\mathrm{CO_2} + 3\mathrm{H_2O} \][/tex]
So, the final coefficients are:
[tex]\[ a = 1, \quad b = 3, \quad c = 2, \quad d = 3 \][/tex]
