Question

Bromine is prepared commercially by the reaction: $2Br^{-}_{\left(aq\right)}+Cl_{2\left(aq\right)} \rightarrow 2Cl^{-}_{\left(aq\right)}+Br_{2\left(aq\right)}$ Suppose we have $50.0\, mL$ of $0.060\, M$ solution of $NaBr.$ What volume of $0.05\, M$ solution of $Cl_2$ is needed to react completely with the $Br^-$ ?

• A. $30\, mL$
• B. $40\, mL$
• C. $20\, mL$
• D. $60\, mL$

Answer 1

Answer: (A)

$50\, mL$ of $0.060 \,M \,NaBr$ contain $NaBr$
$=\frac{0.060}{1000}\times50\,mol=0.003\,mol$
$2 \,mol$ of $Br^{-}$ react with $Cl_{2} = 1\, mol$
$\therefore 0.00_{3}\, mol$ of $Br^{-} $will react with $Cl_{2}=\frac{1}{2}\times 0.003$
$=0.0015\, mol$
$0.05\, mol\, Cl_{2}$ solution are present in $1000 \,mL$ of $Cl_{2} $solution
$\therefore 0.0015\, mol$ of $Cl_{2}$ will be present in $Cl_{2}$ solution
$=\frac{1000}{0.05}\times0.0015=30\,mL $