\(n_{O_2} = \dfrac{5,6}{22,4} = 0,25(mol)\\ 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ n_{KClO_3\ pư} = \dfrac{2}{3}n_{O_2} = \dfrac{1}{6}(mol)\\ n_{KClO_3\ đã\ dùng} = \dfrac{ \dfrac{1}{6}}{85\%} = \dfrac{10}{51}(mol)\\ m_{KClO_3} = \dfrac{10}{51}.122,5 = \dfrac{1225}{51} = 24,02(gam)\)