4)

Phản ứng xảy ra:

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

Ta có: 

\({n_{KMn{O_4}}} = \frac{{1,58}}{{39 + 55 + 16.4}} = 0,01{\text{ mol}} \to {{\text{n}}_{{O_2}}} = \frac{1}{2}{n_{KMn{O_4}}} = 0,005{\text{ mol}} \to {\text{V = }}{{\text{V}}_{{O_2}}} = 0,005.22,4 = 0,112{\text{ lít}}\)

\(2Zn + {O_2}\xrightarrow{{}}2ZnO\)

Ta có: \({n_{Zn}} = \frac{{0,975}}{{65}} = 0,015{\text{ mol > 2}}{{\text{n}}_{{O_2}}}\) nên Zn dư.

\( \to {m_{rắn}} = {m_{Zn}} + {m_{{O_2}}} = 0,975 + 0,005.32 = 1,135{\text{ gam}}\)

5)

Phản ứng xảy ra:

\(4R + 3{O_2}\xrightarrow{{{t^o}}}2{R_2}{O_3}\)

Ta có: 

\({n_R} = 2{n_{{R_2}{O_3}}} \to \frac{{2,16}}{R} = 2.\frac{{4,08}}{{2R + 16.3}} \to R = 27 \to Al\) (nhôm)