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 involves phenol (C6H5OH) reacting with bromine (Br2) to form tribromophenol (C6H2Br3OH). The balanced chemical equation can be represented as:

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

### Step 2: Determine the Moles of Bromine Required
From the balanced equation, we can see that one mole of phenol reacts with three moles of bromine. Therefore, to produce tribromophenol from one mole of phenol, we need:

\[ 3 \text{ moles of } \text{Br}_2 \]

### Step 3: Convert Moles of Bromine to Grams
To find out how much bromine is needed in grams, we need to calculate the mass of 3 moles of bromine (Br2). The molar mass of bromine (Br) is approximately 80 g/mol, so the molar mass of bromine gas (Br2) is:

\[ 2 \times 80 \, \text{g/mol} = 160 \, \text{g/mol} \]

Now, for 3 moles of Br2:

\[ \text{Mass of } 3 \text{ moles of } \text{Br}_2 = 3 \, \text{moles} \times 160 \, \text{g/mol} = 480 \, \text{grams} \]

### Conclusion
Therefore, to produce tribromophenol from one mole of phenol, we need 3 moles of bromine, which is equivalent to 480 grams of bromine.

### Final Answer
- **3 moles of bromine (Br2) are needed.**
- **480 grams of bromine (Br2) are needed.**

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