$n_B=\dfrac{0,48}{B}(mol)$
$B+2HCl\to BCl_2+H_2$
$\Rightarrow n_{BCl_2}=n_{H_2}=\dfrac{0,48}{B}(mol)$
$m_{dd\text{spứ}}=m_B+200-m_{H_2}$
$=0,48+200-\dfrac{0,48.2}{B}=\dfrac{200,48B-0,96}{B}(g)$
$m_{BCl_2}=\dfrac{0,48(B+71)}{B}(g)$
$\Rightarrow \dfrac{0,48B+34,08}{B}=0,219.\dfrac{200,48-0,96B}{B}$
$\Leftrightarrow 0,48B+34,08=43,90512-0,21024B$
$\Leftrightarrow 0,69024B=9,82512$
$\Leftrightarrow B=14$ (N không là kim loại, vô lí)