Đáp án:
a, \({V_{{C_2}{H_4}}} = 0,448l\)
b, \({V_{Ca{{(OH)}_2}}} = \dfrac{n}{{CM}} = 0,03l\)
Giải thích các bước giải:
\(\begin{array}{l}
{C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O\\
{V_{{O_2}}} = \dfrac{1}{5}{V_{KK}} = 1,344l \to {n_{{O_2}}} = 0,06mol\\
\to {n_{{C_2}{H_4}}} = \dfrac{1}{3}{n_{{O_2}}} = 0,02mol \to {V_{{C_2}{H_4}}} = 0,448l\\
Ca{(OH)_2} + C{O_2} \to CaC{O_3} + {H_2}O\\
{n_{C{O_2}}} = \dfrac{2}{3}{n_{{O_2}}} = 0,04mol\\
\to {n_{Ca{{(OH)}_2}}} = {n_{C{O_2}}} = 0,04mol\\
\to {V_{Ca{{(OH)}_2}}} = \dfrac{n}{{CM}} = 0,03l
\end{array}\)
