Em tham khảo nha :

\(\begin{array}{l}
2KMn{O_4} \to Mn{O_2} + {O_2} + {K_2}Mn{O_4}(1)\\
2KCl{O_3} \to 2KCl + 3{O_2}(2)\\
a)\\
{n_{{O_2}}} = \dfrac{m}{M} = \dfrac{{3,2}}{{32}} = 0,1mol\\
{n_{KMn{O_4}}} = 2{n_{{O_2}}} = 0,2mol\\
{m_{KMn{O_4}}} = n \times M = 0,2 \times 158 = 31,6g\\
{n_{KCl{O_3}}} = \dfrac{2}{3}{n_{{O_2}}} = \dfrac{1}{{15}}mol\\
{m_{KCl{O_3}}} = n \times M = \dfrac{1}{{15}} \times 123 = 8,2g\\
b)\\
{n_{KMn{O_4}}} = 0,1mol\\
{n_{{O_2}(1)}} = \dfrac{{{n_{KMn{O_4}}}}}{2} = 0,05mol\\
{m_{{O_2}(1)}} = n \times M = 0,05 \times 32 = 1,6g\\
{n_{KCl{O_3}}} = 0,1mol\\
{n_{{O_2}}} = \dfrac{3}{2}{n_{KCl{O_3}}} = 0,15mol\\
{m_{{O_2}}} = n \times M = 0,15 \times 32 = 4,8g\\
{n_{KMn{O_4}}} = \dfrac{m}{M} = \dfrac{{50}}{{158}} = \frac{{25}}{{79}}mol\\
{n_{{O_2}(1)}} = \dfrac{{{n_{KMn{O_4}}}}}{2} = \dfrac{{25}}{{158}}mol\\
{m_{{O_2}(1)}} = n \times M = \dfrac{{25}}{{158}} \times 32 = 5,063g\\
{n_{KCl{O_3}}} = \dfrac{m}{M} = \dfrac{{50}}{{123}}mol\\
{n_{{O_2}(2)}} = \dfrac{3}{2}{n_{KCl{O_3}}} = \dfrac{{25}}{{41}}mol\\
{m_{{O_2}(2)}} = n \times M = \dfrac{{25}}{{41}} \times 32 = 19,512g
\end{array}\)

 

Làm riêng khá dài, mình gộp luôn nhé

a,

$a_1,$

$2KClO_3\xrightarrow{t^o} 2KCl+3O_2 \ \ (1)$

$2KMnO_4\xrightarrow{t^o} K_2MnO_4+MnO_2+O_2 \ \ (2)$

Số mol $O_2$ cần điều chế là:

$n_{O_2}=\dfrac{3.2}{32}=0.1 \ (mol)$

$Theo \ pt \ (1):n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}·0.1=\dfrac{1}{15} \ (mol)$

$Theo \ pt \ (2):n_{KMnO_4}=2n_{O_2}=2·0.1=0.2 \ (mol)$

$a_2,$

$m_{KClO_3}=\dfrac{1}{15}·122.5=8.167 \ (g)$

$m_{KMnO_4}=0.2·158=31.6 \ (g)$

b,

$b_1,$

$Theo \ pt \ (1):n_{O_2}=\dfrac{3}{2}n_{KClO_3}=\dfrac{3}{2}·0.1=0.15 \ (mol)$

Khối lượng $O_2$ thu được là:

$m_{O_2}=0.15·32=4.8 \ (g)$

$Theo  \ pt \ (2):n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=\dfrac{1}{2}·0.1=0.05 \ (mol)$

Khối lượng $O_2$ thu được là:

$m_{O_2}=0.05·32=1.6 \ (g)$

$b_2,$

Số mol $KClO_3$ là:

$n_{KClO_3}=\dfrac{50}{122.5}=\dfrac{20}{49} \ (mol)$

$Theo \ pt \ (1):n_{O_2}=\dfrac{3}{2}n_{KClO_3}=\dfrac{3}{2}·\dfrac{20}{49}=\dfrac{30}{49} \ (mol)$

$m_{O_2}=\dfrac{30}{49}·32=19.6 \ (g)$

Số mol $KMnO_4$ là:

$n_{KMnO_4}=\dfrac{50}{158}=\dfrac{25}{79} \ (mol)$

$Theo \ pt \ (2):n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=\dfrac{1}{2}·\dfrac{25}{79}=\dfrac{25}{158} \ (mol)$

Khối lượng $O_2$ thu được là:

$m_{O_2}=\dfrac{25}{158}·32=5 \ (g)$