4)

Các phản ứng xảy ra:

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

\(2{C_2}{H_2} + 5{O_2}\xrightarrow{{{t^o}}}4C{O_2} + 2{H_2}O\)

\({C_2}{H_6}O + 3{O_2}\xrightarrow{{{t^o}}}2C{O_2} + 3{H_2}O\)

\({C_{12}}{H_{22}}{O_{11}} + 12{O_2}\xrightarrow{{{t^o}}}12C{O_2} + 11{H_2}O\)

5)

Phản ứng xảy ra:

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

a)

\({n_{{O_2}}} = \frac{{6,4}}{{16.2}} = 0,2{\text{ mol = }}{{\text{n}}_{C{O_2}}}\)

\( \to {m_{C{O_2}}} = 0,2.(12 + 16.2) = 8,8{\text{ gam}}\)

b)

\({n_C} = \frac{6}{{12}} = 0,5{\text{ mol;}}{{\text{n}}_{{O_2}}} = \frac{{19,2}}{{16.2}} = 0,6{\text{ mol > }}{{\text{n}}_C}\)

Vậy \(O_2\) dư

\( \to {n_{C{O_2}}} = {n_C} = 0,5{\text{ mol}}\)

\( \to {m_{C{O_2}}} = 0,5.(12 + 16.2) = 22{\text{ gam}}\)

6)

a)

Phản ứng xảy ra:

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

Ta có:

\({n_P} = \frac{{46,5}}{{31}} = 1,5{\text{ mol}}\)

\( \to {n_{{O_2}}} = \frac{5}{4}{n_P} = 1,875{\text{ mol}}\)

\( \to {m_{{O_2}}} = 1,875.32 = 60{\text{ gam}}\)

b)

Phản ứng xảy ra:

\(4Al + 3{O_2}\xrightarrow{{{t^o}}}2A{l_2}{O_3}\)

Ta có:

\({n_{Al}} = \frac{{67,5}}{{27}} = 2,5{\text{ mol}}\)

\( \to {n_{{O_2}}} = \frac{3}{4}{n_{Al}} = 1,875{\text{ mol}}\)

\( \to {m_{{O_2}}} = 1,875.32 = 60{\text{ gam}}\)

c)

Phản ứng xảy ra:

\(2{H_2} + {O_2}\xrightarrow{{{t^o}}}2{H_2}O\)

Ta có:

\({n_{{H_2}}} = \frac{{33,6}}{{22,4}} = 1,5{\text{ mol}}\)

\( \to {n_{{O_2}}} = \frac{1}{2}{n_{{H_2}}} = 0,75{\text{ mol}}\)

\( \to {m_{{O_2}}} = 0,75.32 = 24{\text{ gam}}\)