What amount of bromine will be required to convert 6 g of phenol into 2,4,6-tribromophenol?

30.63 g

The balanced chemical equation for the reaction between phenol and bromine is:

From the balanced equation, we can see that 1 mole of phenol reacts with 3 moles of bromine. This means that the molar ratio of phenol to bromine is 1:3.

To calculate the amount of bromine required to convert 6 g of phenol, we can use the following steps:

Convert the mass of phenol to moles:

\(n(phenol) = \frac{m(phenol)}{M(phenol)}\)

\(n(phenol) = \frac{6 g}{94.11 g/mol}\)

\(n(phenol) = 0.0638 mol\)

Use the molar ratio to find the moles of bromine required:

\(n(bromine) = n(phenol) × (\frac{3\text{ mol }Br_2}{1\text{ mol phenol}})\)

\(n(bromine) = 0.0638 mol × (\frac{3\text{ mol }Br_2}{1\text{ mol phenol}})\)

\(n(bromine) = 0.1914 mol\)

Convert the moles of bromine to mass:

\(m(bromine) = n(bromine) × M(bromine)\)

\(m(bromine) = 0.1914 mol × 159.807 g/mol\)

\(m(bromine) = 30.63 g\)

Therefore, 30.63 g of bromine is required to convert 6 g of phenol into 2,4,6-tribromophenol. The correct answer is 1. 30.63 g.