Question

help

Answer 1

$n_{H_2} = \dfrac{10,08}{22,4} = 0,45(mol)$
$2B + 6HCl \to 2BCl_3 + 3H_2$
Theo PTHH , $n_{BCl_3} = \dfrac{2}{3}n_{H_2} = 0,3(mol)$
$\Rightarrow M_{BCl_3} = B + 35,5.3 = \dfrac{40,05}{0,3} = 133,5(g/mol)$

$\Rightarrow B = 27(Al)$
Vậy kim loại B là Al