$n_{KMnO_4}=\dfrac{m_{KMnO_4}}{M_{KMnO_4}}=\dfrac{15.8}{158}=0.1 \ (mol)$

Phương trình hóa học:

$2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\\ \ \ \ \ 0.1 \ \ \ \ \ \ \ \ \ \ \ \xrightarrow{\quad \quad \quad \quad \quad} \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.05 \ \ (mol)$

Thể tích khí Oxi thu được là:

$V_{O_2}=n_{O_2}·22.4=0.05·22.4=1.12 \ (l)$

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$n_{KClO_2}=\dfrac{m_{KClO_3}}{M_{KClO_3}}=\dfrac{24.5}{122.5}=0.2 \ (mol)$

Phương trình hóa học:

$2KClO_2\xrightarrow{t^o}2KCl+3O_2\uparrow\\ \ \ \ \ 0.2 \ \ \ \ \xrightarrow{\quad \quad \quad } \ \ \ \ \ \ 0.3 \ \ (mol)$

Thể tích khí Oxi thu được là:

$V_{O_2}=n_{O_2}·22.4=0.3·22.4=6.72 \ (l)$

$1) \\n_{KMnO_4}=\dfrac{15,8}{158}=0,1mol \\PTHH : \\2KMnO_4\overset{t^o}\to K_2MnO_4+MnO_2+O_2 \\Theo\ pt : \\n_{O_2}=\dfrac{1}{2}.n_{KMnO_4}=\dfrac{1}{2}.0,1=0,05mol \\⇒V_{O_2}=0,05.22,4=1,12l \\2) \\n_{KClO_3}=\dfrac{24,5}{122,5}=0,2mol \\PTHH : \\2KClO_3\overset{t^o}\to 2KCl+3O_2 \\Theo\ pt : \\n_{O_2}=\dfrac{3}{2}.n_{KClO_3}=\dfrac{3}{2}.0,2=0,3mol \\⇒V_{O_2}=0,3.22,4=6,72l$