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

Gọi `x` là khối lượng `KMnO_4` phản ứng

`=>m_(KMnO_4  (dư))=22,12-158x`

`m_(K_2MnO_4+MnO_2)=197.0,5x+87.0,5x=142x`

`=>22,12-158x+142x=21,26`

`=>x=0,05375(mol)`

`a,`

`V_(O_2)=0,026875.22,4=0,602(l)`

`b,`

`%_(KMnO_4)=\frac{0,05375}{0,14}.100=38,39%`

`c,`

$2HgO\xrightarrow{t^o}2Hg+O_2$

`0,05375`                         `0,026875`

`=>m_(HgO)=0,05375.217=11,66375(g)`

Đáp án:

\({V_{{O_2}}} = 0,602{\text{ lít}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(2KMn{O_4}\xrightarrow{{{t^o}}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)

BTKL:

\({m_{KMn{O_4}}} = {m_{rắn}} + {m_{{O_2}}}\)

\( \to {m_{{O_2}}} = 22,12 - 21,26 = 0,86{\text{ gam}}\)

\( \to {n_{{O_2}}} = \frac{{0,86}}{{32}} = 0,026875{\text{ mol}}\)

\( \to {V_{{O_2}}} = 0,026875.22,4 = 0,602{\text{ lít}}\)

\({n_{KMn{O_4}}} = \frac{{22,12}}{{39 + 55 + 16.4}} = 0,14{\text{ mol}}\)

\( \to {n_{KMn{O_4}{\text{ phản ứng}}}} = 2{n_{{O_2}}} = 0,05375{\text{ mol}}\)

\( \to \% KMn{O_4}{\text{ phản ứng = }}\frac{{0,05375}}{{0,14}} = 38,39\% \)

\(2HgO\xrightarrow{{{t^o}}}2Hg + {O_2}\)

\( \to {n_{HgO}} = 2{n_{{O_2}}} = 0,05375{\text{ mol}}\)

\( \to {m_{HgO}} = 0,05375.(201 + 16) = 11,66375{\text{ gam}}\)