Đáp án:
\(\begin{array}{l}
a)\\
CTPT:{C_4}{H_8}\\
b)\\
{m_{{C_4}{H_8}}} = 2{m_{{N_2}}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{C_n}{H_{2n}} + \dfrac{{3n}}{2}{O_2} \to nC{O_2} + n{H_2}O\\
{n_{anken}} = \dfrac{{0,4}}{{22,4}} = \frac{1}{{56}}mol\\
{n_{{O_2}}} = \dfrac{{2,4}}{{22,4}} = \frac{3}{{28}}mol\\
{n_{{O_2}}} = \dfrac{{3n}}{2}{n_{anken}}\\
\dfrac{3}{{28}} = \dfrac{{3n}}{2} \times \dfrac{1}{{56}} \Rightarrow n = 4\\
\Rightarrow CTPT:{C_4}{H_8}\\
b)\\
{n_{{C_4}{H_8}}} = {n_{{N_2}}} = \dfrac{1}{{22,4}} = \dfrac{5}{{112}}mol\\
{m_{{C_4}{H_8}}} = \dfrac{5}{{112}} \times 56 = 2,5g\\
{m_{{N_2}}} = \dfrac{5}{{112}} \times 28 = 1,25g\\
\dfrac{{{m_{{C_4}{H_8}}}}}{{{m_{{N_2}}}}} = \dfrac{{2,5}}{{1,25}} = 2\\
{m_{{C_4}{H_8}}} = 2{m_{{N_2}}}
\end{array}\)
