Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{{C_2}{H_4}}} = 30,43\% \\
\% {m_{C{H_4}}} = 69,57\% \\
b)\\
{V_{kk}} = 30,8l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2)\\
a)\\
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
{n_{B{r_2}}} = \dfrac{4}{{160}} = 0,025\,mol\\
{n_{{C_2}{H_4}}} = {n_{B{r_2}}} = 0,025\,mol\\
{n_{C{H_4}}} = \dfrac{{2,8 - 0,025 \times 22,4}}{{22,4}} = 0,1\,mol\\
{m_{{C_2}{H_4}}} = 0,025 \times 28 = 0,7g\\
{m_{C{H_4}}} = 0,1 \times 16 = 1,6g\\
\% {m_{{C_2}{H_4}}} = \dfrac{{0,7}}{{0,7 + 1,6}} \times 100\% = 30,43\% \\
\% {m_{C{H_4}}} = 100 - 30,43 = 69,57\% \\
b)\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
{C_2}{H_4} + 3{O_2} \xrightarrow{t^0} 2C{O_2} + 2{H_2}O\\
{n_{{O_2}}} = 0,1 \times 2 + 0,025 \times 3 = 0,275\,mol\\
{V_{kk}} = 0,275 \times 5 \times 22,4 = 30,8l
\end{array}\)
