$8$g of sulphur is burnt to form $S{O}_2$ which is oxidized by chlorine water. The solution is treated with $BaC{l}_2$ solution. The amount of $BaS{O}_4$ precipitated is

The reactions for the processes are

S + O_2​ ⟶ SO₂ 

SO_2​ + Cl₂​ +2H_2​O ⟶ H_2​SO₄​ + 2HCl

H_2​SO_4​ + BaCl₂​  ⟶  BaSO₄​ + 2HCl​

from the above reactions, 1mole of 'S' produce 1mole of BaSO₄

=$no. of moles of S{O}_2=8/32=0.25$

=$\mathrm{No}. \mathrm{of} \mathrm{moles} \mathrm{of} {\mathrm{BaSO}}_4=0.25\mathrm{moles}$ 

Hence correct option is C.