Đáp án:
$a)$
\[V_{O_2}=0,1.20=2(l)\]
Thể tích O2 cần thu được:
\[V_{O_2}=\dfrac{2}{90\%}=\dfrac{20}{9}(l)\]
\[\to n_{O_2}=\dfrac{\dfrac{20}{9}}{22,4}=\dfrac{25}{252}(mol)\]
Phản ứng xảy ra
\[2KMnO_4\to K_2MnO_4+MnO_2+O_2\]
$\to n_{KMnO_4}=\dfrac{25}{252}.2=\dfrac{25}{126}(mol)$
$\to m_{KMnO_4}=158\times \dfrac{25}{126}=31,35(g)$
b. \[2KClO_3\to 2KCl + 3O_2\]
\[n_{KClO_3}=\dfrac{2}{3}\times n_{O_2}=\dfrac 23. \dfrac{25}{252}=\dfrac{25}{378}(mol)\]
\[\to m_{KClO_3}=122,5\times \dfrac{25}{378}=8,1(g)\]