\(n_{O_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ PTHH:2KClO_3-^{t^o}>2KCl+3O_2\)
ti le 2 : 2 : 3
n(mol) 0,2<---------------0,2<------0,3
\(m_{KClO_3}=n\cdot M=0,2\cdot\left(39+35,5+16\cdot3\right)=24,5\left(g\right)\\ m_{KCl}=n\cdot M=0,2\cdot\left(39+35,5\right)=14,9\left(g\right)\)