Đáp án:
\(\begin{array}{l}
a)\\
\% {V_{{C_2}{H_2}}} = 75\% \\
\% {V_{C{H_4}}} = 25\% \\
b)\\
{V_{{\rm{dd}}B{r_2}}} = 480g\\
c)\\
{V_{kk}} = 106,4l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{C_2}{H_2} + 2B{r_2} \to {C_2}{H_2}B{r_4}\\
{n_{{C_2}{H_2}}} = \dfrac{{7,8}}{{26}} = 0,3\,mol\\
\% {V_{{C_2}{H_2}}} = \dfrac{{0,3 \times 22,4}}{{8,96}} \times 100\% = 75\% \\
\% {V_{C{H_4}}} = 100 - 75 = 25\% \\
b)\\
{n_{B{r_2}}} = 2{n_{{C_2}{H_2}}} = 0,6\,mol\\
{V_{{\rm{dd}}B{r_2}}} = \dfrac{{0,6 \times 160}}{{20\% }} = 480g\\
c)\\
{n_{C{H_4}}} = \dfrac{{8,96}}{{22,4}} - 0,3 = 0,1\,mol\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
2{C_2}{H_2} + 5{O_2} \xrightarrow{t^0} 4C{O_2} + 2{H_2}O\\
{n_{{O_2}}} = 0,1 \times 2 + 0,3 \times \frac{5}{2} = 0,95\,mol\\
{V_{kk}} = 0,95 \times 5 \times 22,4 = 106,4l
\end{array}\)
