Đáp án:
\(\begin{array}{l}
a)\\
{V_{{O_2}}} = 38,08l\\
b)\\
{V_{kk}} = 190,4l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{*{20}{l}}
{a)}\\
{C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O}\\
{2{C_4}{H_{10}} + 13{O_2} \to 8C{O_2} + 10{H_2}O}\\
{{n_{C{H_4}}} = \dfrac{m}{M} = \dfrac{{3,2}}{{16}} = 0,2mol}\\
{{n_{{C_4}{H_{10}}}} = \dfrac{m}{M} = \dfrac{{11,6}}{{58}} = 0,2mol}\\
{{n_{{O_2}}} = 2{n_{C{H_4}}} + \dfrac{{13}}{2}{n_{{C_4}{H_{10}}}} = 2 \times 0,2 + \dfrac{{13}}{2} \times 0,2 = 1,7mol}\\
{{V_{{O_2}}} = n \times 22,4 = 1,7 \times 22,4 = 38,08l}\\
{b)}\\
{{V_{kk}} = \dfrac{{{V_{{O_2}}} \times 100}}{{20}} = \dfrac{{38,08 \times 100}}{{20}} = 190,4l}\\
{}
\end{array}\)
