1)

Phản ứng xảy ra:

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

Ta có:

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

BTKL:

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

\( \to 22,12 = 21,26 + {m_{{O_2}}}\)

\( \to {m_{{O_2}}} = 0,86{\text{ gam}}\)

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

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

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

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

Điều chế \(O_2\)

\(2{H_2}O\xrightarrow{{đp}}2{H_2} + {O_2}\)

\( \to {n_{{H_2}O{\text{ lt}}}} = 2{n_{{O_2}}} = 0,05375 \to {n_{{H_2}O}} = \frac{{0,05375}}{{80\% }} = 0,0671875\)

\( \to {m_{{H_2}O}} = 0,0671875.18 = 1,209375{\text{ gam}}\)

2)

Xét \(M_2O_n\)

Ta có:

\({M_{{M_2}{O_n}}} = 2{M_M} + n{M_O} = 2{M_M} + 16n\)

\( \to \% {m_M} = \frac{{2{M_M}}}{{2{M_M} + 16n}} = 70\%  \to 2{M_M} = 1,4{M_M} + 11,2n\)

\( \to {M_M} = \frac{{56n}}{3}\)

Thỏa mãn giá trị \(n=3 \to M_M=56\)

Vậy kim loại cần tìm là \(Fe\)