a, PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Ta có: \(n_{KCl}=\dfrac{1,49}{74,5}=0,02\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{KCl}=0,03\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,03.24,79=0,7437\left(l\right)\)
b, Theo PT: \(n_{KClO_3\left(TT\right)}=n_{KCl}=0,02\left(mol\right)\)
\(\Rightarrow m_{KClO_3\left(TT\right)}=0,02.122,5=2,45\left(g\right)\)
\(\Rightarrow H=\dfrac{2,45}{3,5}.100\%=70\%\)
2KClO3=>2KCl+3O2
a, nKCl=1,49/74,5=0,02(mol)
=>nO2=0,03(mol)
=>V O2=0,03.22,4=0,672(l)
b, nKClO3=0,02(mol)
mKClO3=0,02.122,5=2,45(g)
H(KClO3)=2,45/3,5.100%=70%