Gọi a, b là số mol C3H6, C2H2.
$n_{Br_2}= \frac{8}{160}=0,05 mol$
Bảo toàn $\pi$: $a+2b=0,05$ (1)
$C_2H_2+2AgNO_3+2NH_3\to Ag_2C_2\downarrow + 2NH_4NO_3$
$n_{Ag_2C_2}=\frac{4,8}{240}=0,02 mol$
$\Rightarrow b=0,02$ (2)
(1)(2) $\Rightarrow a=0,01; b=0,02$
$C_3H_6+\frac{9}{2}O_2\to 3CO_2+3H_2O$
$C_2H_2+\frac{5}{2}O_2\to 2CO_2+H_2O$
$\Rightarrow n_{O_2}=\frac{9}{2}.0,01+\frac{5}{2}.0,02=0,095 mol$
$\Rightarrow V_{O_2}=0,095.22,4=2,128l$
=> chọn D