1)
Phản ứng xảy ra:
\({C_2}{H_5}OH + 3{O_2}\xrightarrow{{{t^o}}}2C{O_2} + 3{H_2}O\)
Ta có:
\({n_{{C_2}{H_5}OH}} = \frac{{4,6}}{{12.2 + 5 + 17}} = 0,1{\text{ mol}}\)
\( \to {n_{{O_2}}} = 3{n_{{C_2}{H_5}OH}} = 0,1.3 = 0,3{\text{ mol}}\)
\( \to {V_{{O_2}}} = 0,3.22,4 = 6,72{\text{ lít}}\)
\( \to {V_{kk}} = 5{V_{{O_2}}} = 6,72.5 = 33,6{\text{ lít}}\)
2)
Phản ứng xảy ra:
\(2{C_2}{H_2} + 5{O_2}\xrightarrow{{{t^o}}}4C{O_2} + 2{H_2}O\)
Ta có:
\({V_{{O_2}}} = \frac{5}{2}{V_{{C_2}{H_2}}} = 22,4.\frac{5}{2} = 56{\text{ lít}}\)
\( \to {V_{kk}} = 5{V_{{O_2}}} = 56.5 = 280{\text{ lít}}\)