a)2KClO3------>2KCl+3O2
b) n O2=5,6/22,4=0,25(mol)
Theo pthh
n KClO3=2/3n O2=0,1667(mol)
m KClO3=0,1667.122,5=20,42(g)
n KCl=2nO2=0,1667(mol)
m KCl=0,166.74,5=12,42(g)
\(a,PTHH:2KClO_3\rightarrow2KCl+3O_2\)
\(b,n_{O_2}=\frac{m_{O_2}}{22,4}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow n_{KClO_3}=0,25.\frac{2}{3}=\frac{0,5}{3}\left(mol\right)\)
\(n_{KClO_3}=\frac{0,5}{3}.112,5=20,42\left(g\right)\)
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