`C_(6)H_(5)OH(g)+O_(2)rarrCO_(2)(g) +H_(2)O(l)`
Magnitude of volume change if `30ml` of `C_(6)H_(5)OH (g)` is burnt with excess amount of oxgen, is

Options

A. `30 ml`
B. `60 ml`
C. `20 ml`
D. `10 ml`

Explanation

To solve the problem of determining the magnitude of volume change when 30 mL of C₆H₅OH (g) is burnt with excess oxygen, we will follow these steps:

### Step 1: Write the balanced chemical equation.
The combustion of phenol (C₆H₅OH) can be represented as:
\[ C₆H₅OH(g) + O₂(g) \rightarrow CO₂(g) + H₂O(l) \]

To balance the equation, we need to ensure that the number of atoms of each element is equal on both sides.

After balancing, the equation becomes:
\[ C₆H₅OH(g) + 7 O₂(g) \rightarrow 6 CO₂(g) + 3 H₂O(l) \]

### Step 2: Determine the volume of reactants.
Given that we have 30 mL of C₆H₅OH, we can find the volume of oxygen required for the reaction. According to the balanced equation:
- 1 mole of C₆H₅OH reacts with 7 moles of O₂.

Thus, the volume of O₂ required can be calculated as:
\[ \text{Volume of } O₂ = 7 \times \text{Volume of } C₆H₅OH = 7 \times 30 \text{ mL} = 210 \text{ mL} \]

### Step 3: Calculate the total volume of reactants.
The total volume of reactants is the sum of the volumes of C₆H₅OH and O₂:
\[ \text{Total Volume of Reactants} = \text{Volume of } C₆H₅OH + \text{Volume of } O₂ \]
\[ = 30 \text{ mL} + 210 \text{ mL} = 240 \text{ mL} \]

### Step 4: Determine the volume of products.
According to the balanced equation:
- 1 mole of C₆H₅OH produces 6 moles of CO₂ and 3 moles of H₂O.

The volume of products can be calculated as:
\[ \text{Volume of } CO₂ = 6 \times 30 \text{ mL} = 180 \text{ mL} \]
\[ \text{Volume of } H₂O = 3 \times 30 \text{ mL} = 90 \text{ mL} \]

Since water is in the liquid state, we only consider the gaseous products (CO₂) for volume change:
\[ \text{Total Volume of Products} = 180 \text{ mL} \]

### Step 5: Calculate the volume change.
The volume change can be calculated by subtracting the total volume of reactants from the total volume of products:
\[ \text{Volume Change} = \text{Total Volume of Products} - \text{Total Volume of Reactants} \]
\[ = 180 \text{ mL} - 240 \text{ mL} = -60 \text{ mL} \]

### Step 6: Determine the magnitude of volume change.
The magnitude of volume change is the absolute value of the volume change:
\[ \text{Magnitude of Volume Change} = | -60 \text{ mL} | = 60 \text{ mL} \]

### Final Answer:
The magnitude of volume change when 30 mL of C₆H₅OH is burnt with excess oxygen is **60 mL**.

---