Answer :
To balance the combustion reaction between octane ([tex]\(C_8H_{18}\)[/tex]) and oxygen ([tex]\(O_2\)[/tex]), we need to ensure that the number of atoms for each element is the same on both sides of the equation. Let's go through the process step-by-step.
### Step 1: Write the unbalanced equation
[tex]\[2 C_8H_{18} + O_2 \rightarrow CO_2 + H_2O\][/tex]
### Step 2: Count the atoms of each element in octane
For [tex]\(2 C_8H_{18}\)[/tex]:
- Carbon: [tex]\(2 \times 8 = 16\)[/tex]
- Hydrogen: [tex]\(2 \times 18 = 36\)[/tex]
### Step 3: Balance the carbon atoms by adjusting the coefficient of [tex]\(CO_2\)[/tex]
Since we have 16 carbon atoms from octane, we need 16 [tex]\(CO_2\)[/tex] molecules because each [tex]\(CO_2\)[/tex] molecule contains one carbon atom.
[tex]\[2 C_8H_{18} + O_2 \rightarrow 16 CO_2 + H_2O\][/tex]
### Step 4: Balance the hydrogen atoms by adjusting the coefficient of [tex]\(H_2O\)[/tex]
Since we have 36 hydrogen atoms from octane, we need 18 [tex]\(H_2O\)[/tex] molecules because each [tex]\(H_2O\)[/tex] molecule contains two hydrogen atoms.
[tex]\[2 C_8H_{18} + O_2 \rightarrow 16 CO_2 + 18 H_2O\][/tex]
### Step 5: Balance the oxygen atoms
- We have [tex]\(16 CO_2\)[/tex] molecules, contributing [tex]\(16 \times 2 = 32\)[/tex] oxygen atoms.
- We have [tex]\(18 H_2O\)[/tex] molecules, contributing [tex]\(18 \times 1 = 18\)[/tex] oxygen atoms.
Total oxygen atoms required on the product side:
[tex]\[32 + 18 = 50\][/tex]
Since each [tex]\(O_2\)[/tex] molecule contains 2 oxygen atoms, the number of [tex]\(O_2\)[/tex] molecules needed on the reactant side:
[tex]\[ \frac{50}{2} = 25\][/tex]
### Step 6: Write the balanced equation
[tex]\[2 C_8H_{18} + 25 O_2 \rightarrow 16 CO_2 + 18 H_2O\][/tex]
### Final balanced equation
The combustion reaction between octane and oxygen is balanced as follows:
[tex]\[2 C_8H_{18} + 25 O_2 \rightarrow 16 CO_2 + 18 H_2O\][/tex]
### Step 1: Write the unbalanced equation
[tex]\[2 C_8H_{18} + O_2 \rightarrow CO_2 + H_2O\][/tex]
### Step 2: Count the atoms of each element in octane
For [tex]\(2 C_8H_{18}\)[/tex]:
- Carbon: [tex]\(2 \times 8 = 16\)[/tex]
- Hydrogen: [tex]\(2 \times 18 = 36\)[/tex]
### Step 3: Balance the carbon atoms by adjusting the coefficient of [tex]\(CO_2\)[/tex]
Since we have 16 carbon atoms from octane, we need 16 [tex]\(CO_2\)[/tex] molecules because each [tex]\(CO_2\)[/tex] molecule contains one carbon atom.
[tex]\[2 C_8H_{18} + O_2 \rightarrow 16 CO_2 + H_2O\][/tex]
### Step 4: Balance the hydrogen atoms by adjusting the coefficient of [tex]\(H_2O\)[/tex]
Since we have 36 hydrogen atoms from octane, we need 18 [tex]\(H_2O\)[/tex] molecules because each [tex]\(H_2O\)[/tex] molecule contains two hydrogen atoms.
[tex]\[2 C_8H_{18} + O_2 \rightarrow 16 CO_2 + 18 H_2O\][/tex]
### Step 5: Balance the oxygen atoms
- We have [tex]\(16 CO_2\)[/tex] molecules, contributing [tex]\(16 \times 2 = 32\)[/tex] oxygen atoms.
- We have [tex]\(18 H_2O\)[/tex] molecules, contributing [tex]\(18 \times 1 = 18\)[/tex] oxygen atoms.
Total oxygen atoms required on the product side:
[tex]\[32 + 18 = 50\][/tex]
Since each [tex]\(O_2\)[/tex] molecule contains 2 oxygen atoms, the number of [tex]\(O_2\)[/tex] molecules needed on the reactant side:
[tex]\[ \frac{50}{2} = 25\][/tex]
### Step 6: Write the balanced equation
[tex]\[2 C_8H_{18} + 25 O_2 \rightarrow 16 CO_2 + 18 H_2O\][/tex]
### Final balanced equation
The combustion reaction between octane and oxygen is balanced as follows:
[tex]\[2 C_8H_{18} + 25 O_2 \rightarrow 16 CO_2 + 18 H_2O\][/tex]
