PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
a, Ta có: \(n_{O_2}=\dfrac{48}{32}=1,5\left(mol\right)\)
Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=1\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=1.122,5=122,5\left(g\right)\)
b, Ta có: \(n_{O_2}=\dfrac{123,93}{22,4}\approx5,53\left(mol\right)\)
Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=3,69\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=3,69.122,5=452,025\left(g\right)\)