How many litres of ${\mathrm{Cl}}_2$ at STP will be liberated by oxidation of NaCl with 10 g ${\mathrm{KMnO}}_4$ ?

  $1\mathrm{gm} \mathrm{eqwt} \mathrm{of} {\mathrm{KMnO}}_{4 } \mathrm{will} \mathrm{liberate} 1 \mathrm{gm} \mathrm{eqwt} \mathrm{of} {\mathrm{Cl}}_2$

 $\frac{158}{5}\mathrm{gm} \mathrm{of} {\mathrm{KMnO}}_{4 }\mathrm{will} \mathrm{liberate} 35.5 \mathrm{gm} \mathrm{of} {\mathrm{Cl}}_2 \mathrm{or} 11.2 \mathrm{lts} \mathrm{of} {\mathrm{Cl}}_2$

Thus 10 gm 0f KMnO₄  will liberate      $\frac{10\times 11.2\times 5}{158}=3.54\mathrm{lts}$