1)
Phản ứng xảy ra:
\(2R + {O_2}\xrightarrow{{{t^o}}}2RO\)
Ta có:
\({n_{{O_2}}} = \frac{{1,344}}{{22,4}} = 0,06{\text{ mol}} \to {{\text{n}}_R} = 2{n_{{O_2}}} = 0,06.2 = 0,12{\text{ mol}}\)
\( \to {M_R} = \frac{{2,688}}{{0,12}} = 24 \to R:Mg\) (magie)
2)
Phản ứng xảy ra:
\(4R + {O_2}\xrightarrow{{{t^o}}}2{R_2}O\)
Ta có:
\({n_{{O_2}}} = \frac{{0,672}}{{22,4}} = 0,03{\text{ mol}} \to {{\text{n}}_R} = 4{n_{{O_2}}} = 0,03.4 = 0,12{\text{ mol}}\)
\( \to {M_R} = \frac{{2,76}}{{0,12}} = 23 \to R:Na\)
\({n_{N{a_2}O}} = \frac{1}{2}{n_{Na}} = 0,06{\text{ mol}}\)
\( \to {m_{N{a_2}O}} = 0,06.(23.2 + 16) = 3,72{\text{ gam}}\)
3)
Phản ứng xảy ra:
\(4R + n{O_2}\xrightarrow{{{t^o}}}2{R_2}{O_n}\)
BTKL:
\({m_R} + {m_{{O_2}}} = {m_{{R_2}{O_n}}}\)
\( \to 14,08 + {m_{{O_2}}} = 17,6 \to {m_{{O_2}}} = 3,52{\text{ gam}} \to {{\text{n}}_{{O_2}}} = \frac{{3,52}}{{32}} = 0,11{\text{ mol}}\)
\( \to {n_R} = \frac{{4{n_{{O_2}}}}}{n} = \frac{{0,11.4}}{n} = \frac{{0,44}}{n}\)
\( \to {M_R} = \frac{{14,08}}{{\frac{{0,44}}{n}}} = 32n\)
Thỏa mãn \(n=2 \to M_R=64 \to R:Cu\) (đồng)
