\(m_{KClO_3\left(pư\right)}=\dfrac{15.90}{100}=13,5\left(g\right)\)
=> \(n_{KClO_3\left(pư\right)}=\dfrac{13,5}{122,5}=\dfrac{27}{245}\left(mol\right)\)
PTHH: 2KClO3 --to--> 2KCl + 3O2
\(\dfrac{27}{245}\)----------------->\(\dfrac{81}{490}\)
=> \(V_{O_2}=\dfrac{81}{490}.22,4=\dfrac{648}{175}\left(l\right)\)
\(n_{KClO_3}=\dfrac{m_{KClO_3}}{M_{KClO_3}}=\dfrac{15}{122,5}=\dfrac{6}{49}mol\)
\(n_{KClO_3}=\dfrac{6}{49}:90\%=\dfrac{20}{147}mol\)
\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)
20/147 10/49 ( mol )
\(V_{O_2}=n_{O_2}.22,4=\dfrac{10}{49}.22,4=4,5714l\)
