\(m_{giảm}=m_{O_2}=13,44\left(g\right)\\ \Rightarrow n_{O_2}=\dfrac{13,44}{32}=0,42\left(mol\right)\\ n_{KClO_3}=\dfrac{49}{122,4}=0,4\left(mol\right)\)
PTHH: \(2KClO_3\xrightarrow[]{t^o}2KCl+3O_2\)
0,28<------------------0,42
\(\Rightarrow H=\dfrac{0,28}{0,4}.100\%=70\%\)
\(n_{KClO_3}=\dfrac{49}{122,5}=0,4\left(mol\right)\)
\(m_{giảm}=m_{O_2}=13,44\left(g\right)\)
\(n_{O_2}=\dfrac{13,44}{32}=0,42\left(mol\right)\)
\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\uparrow\)
bđ: 0,4 0,42 ( mol )
lt: 0,4 0,6 ( mol )
\(H=\dfrac{0,42}{0,6}.100=70\%\)
