$n_{KClO_3\ lt}=\dfrac{12,25}{122,5}=0,1mol \\n_{O_2}=\dfrac{2,24}{22,4}=0,1mol \\PTHH : \\2KClO_3\overset{t^o}\to 2KCl+3O_2 \\Theo\ pt : \\n_{KClO_3\ tt}=\dfrac{2}{3}.n_{O_2}=\dfrac{2}{3}.0,2=\dfrac{1}{15}(mol) \\⇒H=\dfrac{\dfrac{1}{15}}{0,1}=66,67\%$$