\(2KClO_3\xrightarrow[]{t^o}2KCl+3O_2\)
a) Ta có : \(n_{KCl}=\dfrac{14,9}{74,5}=0,2\left(mol\right)\)
Theo Pt : \(n_{KCl}=n_{KClO3}=0,2\left(mol\right)\Rightarrow m_{KClO3}=0,2.122,5=24,5\left(g\right)\)
b) Ta có : \(n_{KCl}=\dfrac{2,98}{74,5}=0,04\left(mol\right)\)
Theo Pt : \(n_{KCl}=n_{KClO3}=0,04\left(mol\right)\Rightarrow m_{KClO3}=0,04.122,5=4,9\left(g\right)\)
c) Theo Pt : \(n_{O2}=\dfrac{3}{2}n_{KClO3}=3,6\left(mol\right)\Rightarrow V_{O2\left(dkc\right)}=3,6.24,79=89,244\left(l\right)\)
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