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 determine the amount of bromine required to convert 2 g of phenol into 2,4,6-tribromophenol, we need to consider the balanced chemical reaction and the stoichiometry involved. Let's go through the steps: 

1. The reaction between phenol \( \text{C}_6\text{H}_5\text{OH} \) and bromine (\( \text{Br}_2 \)) to form 2,4,6-tribromophenol \( \text{C}_6\text{H}_2\text{Br}_3\text{OH} \) is as follows: 
   \(\text{C}_6\text{H}_5\text{OH} + 3\text{Br}_2 \rightarrow \text{C}_6\text{H}_2\text{Br}_3\text{OH} + 3\text{HBr}\)
2. Determine the molar masses:
   • Molar mass of phenol (\( \text{C}_6\text{H}_5\text{OH} \)): 
     Carbon (C): 12 × 6 = 72 g/mol 
     Hydrogen (H): 1 × 6 = 6 g/mol 
     Oxygen (O): 16 × 1 = 16 g/mol 
     Total = 72 + 6 + 16 = 94 g/mol
   • Molar mass of bromine (\( \text{Br}_2 \)): 
     Bromine (Br): 80 × 2 = 160 g/mol
3. Determine the moles of phenol: 
   \(n_{\text{phenol}} = \frac{2 \text{ g}}{94 \text{ g/mol}} \approx 0.0213 \text{ mol}\)
4. According to the balanced equation, 1 mole of phenol reacts with 3 moles of bromine. Therefore, the moles of bromine required will be: 
   \(n_{\text{Br}_2} = 3 \times 0.0213 = 0.0639 \text{ mol}\)
5. Calculate the mass of bromine needed: 
   \(m_{\text{Br}_2} = 0.0639 \text{ mol} \times 160 \text{ g/mol} = 10.224 \text{ g}\)

Therefore, the amount of bromine required is approximately 10.22 g, corresponding to the correct option: 10.22 g. This calculation confirms that option \(10.22 \text{ g}\) is correct.

Answer 2

To determine the amount of bromine needed to convert 2 g of phenol into 2,4,6-tribromophenol, we follow this procedure:

Step 1: Calculate the molar mass of phenol (C₆H₅OH)

Molar mass of C = 12 g/mol
Molar mass of H = 1 g/mol
Molar mass of O = 16 g/mol

Molar mass of phenol = 6(12) + 6(1) + 16 = 94 g/mol

Step 2: Calculate the number of moles of phenol

Number of moles = mass / molar mass = 2 g / 94 g/mol ≈ 0.0213 mol

Step 3: Determine the reaction and molar mass of tribromophenol

Reaction: C₆H₅OH + 3 Br₂ → C₆H₂Br₃OH + 3 HBr

Molar mass of Br₂ = 2(80) = 160 g/mol

Step 4: Calculate bromine required for the reaction

Bromine required per mol of phenol = 3 mols of Br₂
Number of moles of Br₂ = 3 × 0.0213 mol = 0.0639 mol
Mass of Br₂ = moles × molar mass = 0.0639 mol × 160 g/mol = 10.224 g

Thus, the amount of bromine required is 10.22 g.