2)

a) \(C + {O_2}\xrightarrow{{{t^o}}}C{O_2}\)

b) \(3Fe + 2{O_2}\xrightarrow{{{t^o}}}F{e_3}{O_4}\)

c) \(4P + 5{O_2}\xrightarrow{{{t^o}}}2{P_2}{O_5}\)

d)\(2{H_2}O\xrightarrow{{dp}}2{H_2} + {O_2}\)

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

f) \({C_2}{H_4} + 3{O_2}\xrightarrow{{{t^o}}}2C{O_2} + 2{H_2}O\)

3)

Phản ứng xảy ra:

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

Ta có:

\({n_{KMn{O_4}}} = \frac{{15,8}}{{39 + 55 + 16.4}} = 0,1{\text{ mol}}\)

\( \to {n_{{O_2}}} = \frac{1}{2}{n_{KMn{O_4}}} = 0,05{\text{ mol}}\)

\( \to {V_{{O_2}}} = 0,05.22,4 = 1,12{\text{ lít}}\)

4)

Phản ứng xảy ra:

\(S + {O_2}\xrightarrow{{{t^o}}}S{O_2}\)

Ta có:

\({n_S} = \frac{{3,2}}{{32}} = 0,1{\text{ mol;}}{{\text{n}}_{{O_2}}} = \frac{{2,688}}{{22,4}} = 0,12{\text{ mol > }}{{\text{n}}_S}\)

Vậy \(O_2\) dư

\( \to {n_{{O_2}{\text{ dư}}}} = 0,12 - 0,1 = 0,02{\text{ mol}}\)

\( \to {V_{{O_2}{\text{ dư}}}} = 0,02.22,4 = 0,448{\text{ lít}}\)

\({n_{S{O_2}}} = {n_S} = 0,1{\text{ mol}}\)

\( \to {V_{S{O_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)

Đáp án:

 

Giải thích các bước giải:

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