Question

Reaction, $2Br^{-}(aq)+Cl_{2}(aq)\longrightarrow 2Cl^{-}(aq) +Br_{2}(aq)$, is used for commercial preparation of bromine from its salts. Suppose we have 50.0 ml of a 0.060M solution of NaBr. What volume of a 0.050 M solution of $Cl_{2}$ is needed to react completely with $Br^{-}$?

Answer 1

"The given chemical reaction is as follows

$2{ Br }^{ - }(aq)\quad +\quad { Cl }_{ 2 }(aq)\quad \rightarrow \quad 2{ Cl }^{ - }(aq)\quad +\quad Br_{ 2 }$

From the given,

Volume of solution = 50 ml

Molarity of NaBr solution = 0.06 M

From 50 ml and 0.06 M of NaBr moles of Br

$\Rightarrow \quad \frac { 0.06\quad \times \quad 50 }{ 1000 } \quad =\quad 0.003\quad moles$

Now, from 0.003 moles of $B{ r }^{ - }$

We get, the moles of ${ Cl }_{ 2 }$

$\Rightarrow \quad \frac { 0.003 }{ 2 } \quad =\quad 0.0015\quad moles.$

Now volume required would be:

$0.0015\quad =\quad 0.05\quad \times \quad V$

$V\quad \Rightarrow \quad \frac { 0.0015 }{ 0.05 } \quad \Rightarrow \quad 0.03\quad litre\quad \Rightarrow \quad 30\quad ml"$