Giải thích các bước giải:
14,
a,
\(\begin{array}{l}
{C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O\\
{n_{{C_2}{H_4}}} = 0,25mol \to {n_{{O_2}}} = 0,75mol\\
\to {V_{{O_2}}} = 16,8l
\end{array}\)
b, \({V_{KK}} = 5{V_{{O_2}}} = 84l\)
15,
\(\begin{array}{l}
2C{H_3}{\rm{COO}}H + Fe \to {(C{H_3}{\rm{COO)}}_2}Fe + {H_2}\\
{n_{C{H_3}{\rm{COO}}H}} = 0,1mol\\
\to {n_{{{(C{H_3}{\rm{COO)}}}_2}Fe}} = \dfrac{1}{2}{n_{C{H_3}{\rm{COO}}H}} = 0,05mol\\
\to {n_{{H_2}}} = \dfrac{1}{2}{n_{C{H_3}{\rm{COO}}H}} = 0,05mol\\
\to {m_{{{(C{H_3}{\rm{COO)}}}_2}Fe}} = 174 \times 0,05 = 8,7g\\
\to {V_{{H_2}}} = 1,12l
\end{array}\)