\(\begin{matrix} 2KClO_3 &\rightarrow 2KCl&+3O_2\\ 2moles & & 3 moles \end{matrix}\)
2 moles of KClO3 = 2× 122.5=245g
3 moles of O2 =3× 22.4=67.2 dm3
hence , 245 g of potacium chlorate will liberate 67.2 dm3 of oxygen
As , here x gram of KClO3 liberate 6.72 dm3 of oxygen at stp
\(\therefore x={245\times 6.72\over 67.2}=24.5 g\)
required mass of potacium chlorate is 24.5 g