Let's balance the reaction step by step:
The given reaction is:
[tex]\[ \text{CH}_3\text{OH} (l) + \text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O} (g) \][/tex]
We need to balance this chemical equation by ensuring that the number of each type of atom on the reactants side (left) equals the number of each type of atom on the products side (right).
### Step 1: Balance the Carbon atoms
In the reactants, there is 1 carbon atom in CH[tex]\(_3\)[/tex]OH.
In the products, there is 1 carbon atom in CO[tex]\(_2\)[/tex].
Already balanced for Carbon:
[tex]\[ 1 \text{ CH}_3\text{OH} + ? \text{ O}_2 \rightarrow 1 \text{ CO}_2 + ? \text{ H}_2\text{O} \][/tex]
### Step 2: Balance the Hydrogen atoms
In the reactants, there are 4 hydrogen atoms in CH[tex]\(_3\)[/tex]OH.
In the products, each H[tex]\(_2\)[/tex]O has 2 hydrogen atoms.
To balance:
[tex]\[ 1 \text{ CH}_3\text{OH} + ? \text{ O}_2 \rightarrow 1 \text{ CO}_2 + 2 \text{ H}_2\text{O} \][/tex]
Now we have 4 hydrogen atoms on both sides.
### Step 3: Balance the Oxygen atoms
Now we need to check the oxygen atoms:
- In the reactants, we have 1 oxygen atom in CH[tex]\(_3\)[/tex]OH and [tex]\(x\)[/tex] oxygen atoms in O[tex]\(_2\)[/tex]. So a total of [tex]\(1 + 2x\)[/tex] oxygen atoms.
- In the products, we have 2 oxygen atoms from CO[tex]\(_2\)[/tex] and 2 oxygen atoms from 2 H[tex]\(_2\)[/tex]O, giving us a total of 4 oxygen atoms.
Setting up the oxygen balance equation:
[tex]\[ 1 + 2x = 4 \][/tex]
Solving for [tex]\(x\)[/tex] (number of O[tex]\(_2\)[/tex] molecules):
[tex]\[ 2x = 3 \][/tex]
[tex]\[ x = \frac{3}{2} \][/tex]
The balanced equation is:
[tex]\[ 1 \text{ CH}_3\text{OH} + \frac{3}{2} \text{ O}_2 \rightarrow 1 \text{ CO}_2 + 2 \text{ H}_2\text{O} \][/tex]
To get integer coefficients, multiply all coefficients by 2:
[tex]\[ 2 \text{ CH}_3\text{OH} + 3 \text{ O}_2 \rightarrow 2 \text{ CO}_2 + 4 \text{ H}_2\text{O} \][/tex]
Now, the equation is balanced with integer coefficients. The coefficient of [tex]\(\text{O}_2\)[/tex] is 3.
