$CH\equiv C-CH_3\buildrel{{Ag^+}}\over\longrightarrow CH_3-C\equiv CAg\downarrow$
$n_{C_3H_4}=n_{\downarrow}=\dfrac{17,64}{12.3+3+108}=0,12(mol)$
Bảo toàn $\pi$: $n_{C_2H_4}+2n_{C_3H_4}=n_{Br_2}$
$\to n_{C_2H_4}=0,34-0,12.2=0,1(mol)$
$\to a=0,1+0,12=0,22(mol)$