Question

The folowing reaction 2Br^{- } + Cl_2 → 2cl^- + Br_2 is used for the commercial preparation of bromine from its salts. suppose we have 50 ml of a 0.080 M solution of NaBr. what volume of a 0.050 M solution of Cl_2

Answer 1

• The reaction is given as:$2B{r}^{-}+C{l}_2⟶2C{l}^{-}+B{r}_2$

• The Milli mol of Br is given by

  $millimolofBr=M\times V\left(ml\right)where,M=molality$

  $=0.80\times 50=4mmol$

• From the balanced equation, we see 2 mol of bromide ions react with 1 mol of chlorine molecule.

• So 4 m mol of bromide ion reacts with 2 m mol of chlorine molecule.

• Hence, the milli mol of chlorine molecule can be calculated as:

  $MillimolofC{l}_2=M_{C{l}_2}\times V$

  $2=0.050\times V$

  $\therefore V=\frac{2}{0.050}=40mL$

• Hence, 40 mL of chlorine molecule of 0.050 M is required to react completely with 50mL of 0.080 M of bromide ions.