PTHH: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Ta có: \(n_{O_2}=\frac{3,2}{32}=0,1\left(mol\right)\)
\(\Rightarrow n_{KClO_3}=\frac{1}{15}mol\) \(\Rightarrow m_{KClO_3}=\frac{1}{15}\cdot122,5\approx8,17\left(g\right)\)
2KClO3-to>2KCl+3O2
0,067--------------------0,1
nO2=3,2\32=0,1 mol
=>mKClO3=0,067.122,5=8,2g