a,
Khí thoát ra là etilen.
$\%V_{C_2H_4}=\dfrac{0,84.100}{3,36}=25\%$
$\to \%V_{C_3H_4}=75\%$
$n_X=\dfrac{3,36}{22,4}=0,15(mol)$
$n_{C_2H_4}=\dfrac{0,84}{22,4}=0,0375(mol)$
$\to n_{C_3H_4}=0,15-0,0375=0,1125(mol)$
Bảo toàn $C$: $n_{C_3H_3Ag\downarrow}=n_{C_3H_4}=0,1125(mol)$
$\to m=0,1125(12.3+3+108)=16,5375g$
b,
$n_{Br_2}=n_{C_2H_4}+2n_{C_3H_4}=0,2625(mol)$
$\to a=0,2625$
