Đáp án:

a, 31g

b, 8g

Giải thích các bước giải:

a, \(\;2KMn{O_4}\; \to {\rm{ }}{K_2}Mn{O_4}\; + Mn{O_2}\; + {O_2}\)

Thể tích oxi cần thu là (20\( \times \)100) + (10% \( \times \)20 \( \times \)100)= 2200(ml)

\(\begin{array}{l}
{n_{{O_2}}} = \dfrac{{11}}{{112}}mol\\
{n_{KMn{O_4}}} = 2{n_{{O_2}}} = \dfrac{{11}}{{56}}mol \to m = 31g
\end{array}\)

b, \(2{\rm{ }}KCl{O_3}{\rm{ }} \to 2KCl{\rm{ }} + 3{\rm{ }}{O_2}\)

\({m_{KCl{O_3}}} = {\rm{ }}\dfrac{{\left( {2200 \times 2 \times 122,5} \right)}}{{\left( {3 \times 22400} \right)}}{\rm{ }} = {\rm{ }}8g\)

a/

Thể tích oxi cần thu :

$(20.100) + (10\%.20.100)= 2200ml=2,2l$

$⇒n_{O_2}=2,2/22,4=\dfrac{11}{112}mol$

$2KMnO_4\overset{t^o}\to K_2MnO_4+MnO_2+O_2$

Theo pt :

$n_{KMnO_4}=2.n_{O_2}=2.\dfrac{11}{112}=\dfrac{11}{56}mol$

$⇒m_{KMnO_4}=\dfrac{11}{56}.158=\dfrac{869}{28}g$

b/

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

Theo pt :

$n_{KClO_3}=2/3.n_{O_2}=2/3.\dfrac{11}{112}=\dfrac{11}{168}mol$

$⇒m_{KClO_3}=\dfrac{11}{168}.122,5=\dfrac{385}{48}g$