Em tham khảo nha:

\(\begin{array}{l}
{P_1}:\\
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
{n_{B{r_2}}} = \dfrac{{80 \times 20\% }}{{160}} = 0,1\,mol\\
{n_{{C_2}{H_4}}} = {n_{B{r_2}}} = 0,1\,mol\\
 \Rightarrow {m_{{C_3}{H_8}}} = 12,2 - 0,1 \times 2 \times 28 = 6,6g\\
{n_{{C_3}{H_8}}} = \dfrac{{6,6}}{{44}} = 0,15\,mol\\
V = (0,1 + 0,15) \times 22,4 = 5,6l\\
\% {V_{{C_2}{H_4}}} = \dfrac{{0,1 \times 22,4}}{{5,6}} \times 100\%  = 40\% \\
\% {V_{{C_3}{H_8}}} = 100 - 40 = 60\% \\
b)\\
{C_3}{H_8} + 5{O_2} \xrightarrow{t^0} 3C{O_2} + 4{H_2}O\\
{C_2}{H_4} + 3{O_2} \xrightarrow{t^0} 2C{O_2} + 2{H_2}O\\
{n_{C{O_2}}} = 0,075 \times 3 + 0,1 \times 2 = 0,425\,mol\\
{n_{{H_2}O}} = 0,075 \times 4 + 0,1 \times 2 = 0,5\,mol\\
C{O_2} + Ba{(OH)_2} \to BaC{O_3} + {H_2}O\\
{n_{BaC{O_3}}} = {n_{C{O_2}}} = 0,425\,mol\\
\Delta m \text{ giảm } = 0,425 \times 197 - 0,425 \times 44 - 0,5 \times 18 = 56,025g
\end{array}\)