\(KClO_3\underrightarrow{t^0}KCl+\dfrac{3}{2}O_2\)(*)
\(a.n_{O_2}=\dfrac{9,6}{32}=0,3mol\)
Theo pt (*) => \(m_{KClO_3}=122,5.\left(0,3\cdot\dfrac{2}{3}\right)=24,5g\)
\(b.n_{O_2}=0,45mol\\ \Rightarrow m_{KClO_3}=122,5\cdot\left(0,45\cdot\dfrac{2}{3}\right)=36,75g\\ c.n_{O_2}=\dfrac{36\cdot10^{21}}{6\cdot10^{23}}=0,06mol\\ \Rightarrow m_{KClO_3}=122,5\cdot\left(0,06\cdot\dfrac{2}{3}\right)=4,9g\)