What amount of bromine will be required to co | vedclass

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(C) The chemical equation for the reaction is:$C_6H_5OH + 3Br_2 \rightarrow C_6H_2Br_3OH + 3HBr$From the stoichiometry of the reaction,$1 \ mole$ of p

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$10.22$

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(C) The chemical equation for the reaction is:$C_6H_5OH + 3Br_2 ightarrow C_6H_2Br_3OH + 3HBr$From the stoichiometry of the reaction,$1 \ mole$ of phenol $(94 \ g/mol)$ reacts with $3 \ moles$ of $Br_2$ $(3 	imes 160 = 480 \ g/mol)$.Therefore,$94 \ g$ of phenol requires $480 \ g$ of $Br_2$.For $2 \ g$ of phenol,the amount of $Br_2$ required is:$	ext{Amount of } Br_2 = \frac{480 \ g}{94 \ g} 	imes 2 \ g \approx 10.2127 \ g \approx 10.22 \ g$.

What amount of bromine will be required to convert $2 \ g$ of phenol into $2, 4, 6-$tribromophenol (in $g$)?

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