\(n_{O_2}=\dfrac{m_{O_2}}{M_{O_2}}=\dfrac{48}{32}=1,5mol\)
\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)
1 1,5 ( mol )
\(m_{KClO_3}=n_{KClO_3}.M_{KClO_3}=1.122,5=122,5g\)
\(n_{O_2}=\dfrac{48}{32}=1,5\left(mol\right)\)
PTHH : 2KClO3 -> 2KCl + 3O2
1 1,5
\(m_{KClO_3}=1.122,5=122,5\left(g\right)\)