`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 calculating the magnitude of volume change when 30 mL of phenol (C₆H₅OH) is burned with excess oxygen, we will follow these steps:

### Step 1: Write the Balanced Chemical Equation
The balanced chemical equation for the combustion of phenol is:
\[ C_6H_5OH(g) + 7O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l) \]

### Step 2: Identify the Volume of Reactants
We are given that the volume of phenol (C₆H₅OH) is 30 mL. Since we have excess oxygen, we will focus on the volume of phenol:
- Volume of C₆H₅OH = 30 mL

### Step 3: Calculate the Volume of Oxygen Required
From the balanced equation, we see that 1 mole of phenol reacts with 7 moles of oxygen. Therefore, the volume of oxygen needed can be calculated as follows:
- Volume of O₂ = 7 × Volume of C₆H₅OH
- Volume of O₂ = 7 × 30 mL = 210 mL

### Step 4: Calculate the Volume of Products Formed
According to the balanced equation, for every 1 mole of phenol, 6 moles of CO₂ and 3 moles of H₂O are produced. We will calculate the volumes of these products:
- Volume of CO₂ = 6 × Volume of C₆H₅OH = 6 × 30 mL = 180 mL
- Volume of H₂O (as a gas) = 3 × Volume of C₆H₅OH = 3 × 30 mL = 90 mL

### Step 5: Calculate Total Volume of Products
The total volume of gaseous products (CO₂ and H₂O) is:
- Total Volume of Products = Volume of CO₂ + Volume of H₂O
- Total Volume of Products = 180 mL + 90 mL = 270 mL

### Step 6: Calculate Total Volume of Reactants
The total volume of reactants (C₆H₅OH and O₂) is:
- Total Volume of Reactants = Volume of C₆H₅OH + Volume of O₂
- Total Volume of Reactants = 30 mL + 210 mL = 240 mL

### Step 7: Calculate the Change in Volume
The change in volume (ΔV) can be calculated as:
\[ \Delta V = \text{Total Volume of Products} - \text{Total Volume of Reactants} \]
\[ \Delta V = 270 mL - 240 mL = 30 mL \]

### Conclusion
The magnitude of the volume change when 30 mL of phenol is burned with excess oxygen is **30 mL**.

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