2)

Phản ứng xảy ra:

\(4P + 5{{\text{O}}_2}\xrightarrow{{{t^o}}}2{P_2}{O_5}\)

Ta có:

\({n_P} = \frac{{12,4}}{{31}} = 0,4{\text{ mol;}}{{\text{n}}_{{O_2}}} = \frac{{13,44}}{{22,4}} = 0,6{\text{ mol > }}\frac{5}{4}{n_P}\)

Vậy \(O_2\) dư

\( \to {n_{{O_2}{\text{ dư}}}} = 0,6 - \frac{5}{4}.0,4 = 0,1{\text{ mol}}\)

\( \to {m_{{O_2}{\text{ dư}}}} = 0,1.32 = 3,2{\text{ gam}}\)

\({n_{{P_2}{O_5}}} = \frac{1}{2}{n_P} = 0,2{\text{ mol}}\)

\( \to {m_{{P_2}{O_5}}} = 0,2.(31.2 + 16.5) = 28,4{\text{ gam}}\)

3)

a)

\(2KN{O_3}\xrightarrow{{{t^o}}}2KN{O_2} + {O_2}\)

Đây là phản ứng phân huỷ

b)

\(2{\text{A}}l + 3C{l_2}\xrightarrow{{{t^o}}}2{\text{A}}lC{l_3}\)

Đây là phản ứng hoá hợp.

4)

Phản ứng xảy ra:

\(3F{\text{e}} + 2{{\text{O}}_2}\xrightarrow{{{t^o}}}F{{\text{e}}_3}{O_4}\)

Ta có:

\({n_{F{\text{e}}}} = \frac{{126}}{{56}} = 2,25{\text{ mol}} \to {{\text{n}}_{{O_2}}} = \frac{2}{3}{n_{F{\text{e}}}} = 1,5{\text{ mol}}\)

\( \to {V_{{O_2}}} = 1,5.22,4 = 33,6{\text{ lít}}\)

Điều chế \(O_2\)

\(2KCl{O_3}\xrightarrow{{{t^o}}}2KCl + 3{{\text{O}}_2}\)

\( \to {n_{KCl{O_3}}} = \frac{2}{3}{n_{{O_2}}} = 1{\text{ mol}}\)

\( \to {m_{KCl{O_3}}} = 1.(39 + 35,5 + 16.3) = 122,5{\text{ gam}}\)