Question

What amount of bromine will be required to convert 2 g of phenol into 2, 4, 6-tribromophenol? (Given molar mass in g mol\(^{-1}\) of C, H, O, Br are 12, 1, 16, 80 respectively)

• A. 10.22 g
• B. 6.0 g
• C. 4.0 g
• D. 20.44 g

Hint:

In stoichiometric calculations, ensure that the moles of the reacting substance are correctly converted to mass using the molar mass. Pay attention to the number of atoms or molecules required for the reaction.

Answer 1

The Correct Option is A

To calculate the bromine needed for the conversion of 2 g of phenol to 2,4,6-tribromophenol, we analyze the balanced chemical equation and stoichiometry. The process is as follows:

1. The reaction between phenol \( \text{C}_6\text{H}_5\text{OH} \) and bromine (\( \text{Br}_2 \)) to yield 2,4,6-tribromophenol \( \text{C}_6\text{H}_2\text{Br}_3\text{OH} \) is:
   \(\text{C}_6\text{H}_5\text{OH} + 3\text{Br}_2 \rightarrow \text{C}_6\text{H}_2\text{Br}_3\text{OH} + 3\text{HBr}\)
2. Molar masses are:
   • Phenol (\( \text{C}_6\text{H}_5\text{OH} \)): 
     C: 12 × 6 = 72 g/mol 
     H: 1 × 6 = 6 g/mol 
     O: 16 × 1 = 16 g/mol 
     Total = 94 g/mol
   • Bromine (\( \text{Br}_2 \)): 
     Br: 80 × 2 = 160 g/mol
3. Moles of phenol are:
   \(n_{\text{phenol}} = \frac{2 \text{ g}}{94 \text{ g/mol}} \approx 0.0213 \text{ mol}\)
4. Based on the stoichiometry, 1 mole of phenol reacts with 3 moles of bromine. Thus, moles of bromine required are:
   \(n_{\text{Br}_2} = 3 \times 0.0213 = 0.0639 \text{ mol}\)
5. The mass of bromine required is:
   \(m_{\text{Br}_2} = 0.0639 \text{ mol} \times 160 \text{ g/mol} = 10.224 \text{ g}\)

Therefore, approximately 10.22 g of bromine is required. This aligns with the correct option: 10.22 g.