$2KClO_3\overset{t^o}{\longrightarrow} 2KCl+3O_2$

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

$\longrightarrow \begin{cases}n_{KCl}=\dfrac{2.0,18}{2}=0,18(mol)\\n_{O_2}=\dfrac{3.0,18}{2}=0,27(mol)\end{cases}$

$\longrightarrow \begin{cases}m_{KCl}=0,18.74,5=13,41(g)\\m_{O_2}=0,27.32=8,64(g)\end{cases}$

Em tham khảo nha :

\(\begin{array}{l}
2KCl{O_3} \to 2KCl + 3{O_2}\\
1)\\
{n_{{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{n_{KCl{O_3}}} = \dfrac{2}{3}{n_{{O_2}}} = 0,2mol\\
{m_{KCl{O_3}}} = n \times M = 0,2 \times 122,5 = 24,5g\\
2)\\
{n_{KCl{O_3}}} = \dfrac{m}{M} = \dfrac{{2,4}}{{122,5}} = 0,02mol\\
{n_{{O_2}}} = \dfrac{3}{2}{n_{KCl{O_3}}} = 0,03mol\\
{n_{KCl}} = {n_{KCl{O_3}}} = 0,02mol\\
3)\\
{n_{KCl{O_3}}} = \dfrac{m}{M} = \dfrac{{22,05}}{{122,5}} = 0,18mol\\
{n_{{O_2}}} = \dfrac{3}{2}{n_{KCl{O_3}}} = 0,27mol\\
{n_{KCl}} = {n_{KCl{O_3}}} = 0,18mol
\end{array}\)