$2KClO_3\overset{t^o}{\longrightarrow} 2KCl+3O_2$
1/ $n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{6,72}{22,4}=0,3(mol)$
Theo pt ta có tỉ lệ: 2:2:3
$\longrightarrow n_{KClO_3}=\dfrac{0,3.2}{3}=0,2(mol)$
$\longrightarrow m_{KClO_3}=n_{KClO_3}.M_{KClO_3}=0,2.122,5=24,5(g)$
2/ Theo pt ta có tỉ lệ: 2:2:3
$\longrightarrow \begin{cases}n_{KCl}=\dfrac{2.2,4}{2}=2,4(mol)\\n_{O_2}=\dfrac{3.2,4}{2}=3,6(mol)\end{cases}$
$\longrightarrow \begin{cases}m_{KCl}=2,4.74,5=178,8(g)\\m_{O_2}=3,6.32=115,2(g)\end{cases}$
3/ $n_{KClO_3}=\dfrac{22,05}{122,5}=0,18(mol)$
Theo pt ta có tỉ lệ: 2:2:3
$\longrightarrow \begin{cases}n_{KCl}=\dfrac{2.0,18}{2}=0,18(mol)\\n_{O_2}=\dfrac{3.0,18}{2}=0,27(mol)\end{cases}$
$\longrightarrow \begin{cases}m_{KCl}=0,18.74,5=13,41(g)\\m_{O_2}=0,27.32=8,64(g)\end{cases}$
