Đáp án:

Bảo toàn khối lượng:

$\rm m_X=m_{ban\ dau}=m_{C_2H_2}+m_{H_2}=0,35×26+0,65×2=10,4\ (g)$

$\rm n_X=\dfrac{10,4}{8×2}=0,65\ (mol)$

$\rm n_{H_2\ (pu)}=n_{ban\ dau}-n_X=0,35+0,65-0,65=0,35\ (mol)$

Bảo toàn nguyên tố C:

$\rm n_{C_2H_2\ (du)}=n_{C_2Ag_2}=\dfrac{24}{240}=0,1\ (mol)$

Bảo toàn liên kết $\pi$:

$\rm 2n_{C_2H_2\ (ban \ dau)}=2n_{C_2H_2\ (du)}+n_{H_2\ (pu)}+n_{Br_2}$

$\rm\to n_{Br_2}=0,15\ (mol)$