How much bromine is needed to produce tribromophenol from one mole of phenol?

Options

A. 3.0 mole
B. 1.5 mole
C. 240 g
D. 480 g

Answer

To determine how much bromine is needed to produce tribromophenol from one mole of phenol, we can follow these steps:

### Step 1: Write the Reaction
The reaction between phenol (C6H5OH) and bromine (Br2) can be represented as follows:

\[ \text{C}_6\text{H}_5\text{OH} + 3 \text{Br}_2 \rightarrow \text{C}_6\text{H}_2\text{Br}_3\text{OH} + 3 \text{HBr} \]

### Step 2: Identify the Stoichiometry
From the balanced reaction, we can see that 1 mole of phenol reacts with 3 moles of bromine to produce tribromophenol.

### Step 3: Calculate the Amount of Bromine Needed
Since we need 3 moles of bromine for 1 mole of phenol, we can conclude that:

- **Bromine needed**: 3 moles

### Step 4: Convert Moles of Bromine to Grams
To find out how much bromine this is in grams, we use the molar mass of bromine. The molar mass of bromine (Br) is approximately 79.9 g/mol. Therefore, for 3 moles:

\[ \text{Mass of Br} = 3 \text{ moles} \times 79.9 \text{ g/mol} = 239.7 \text{ g} \]

### Step 5: Conclusion
Thus, to produce tribromophenol from one mole of phenol, you need:

- **3 moles of bromine** or approximately **239.7 grams of bromine**.

### Final Answer
- **3 moles of bromine** are needed to produce tribromophenol from one mole of phenol.
- In grams, this is approximately **239.7 grams**.