Question

Answer 1

a)

$n_{HCl} = \dfrac{47,45}{36,5} = 1,3(mol)$
$X + 2HCl \to XCl_2 + H_2$

$2Y + 6HCl \to 2YCl_3 + 3H_2$
Theo PTHH : 

$n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,65(mol)$
$V_{H_2} = 0,65.22,4 = 14,56(lít)$

b)

Bảo toàn khối lượng : 

$m_{muối} = m_{hh} + m_{HCl} - m_{H_2} = 12,9 + 47,45 - 0,65.2 = 59,05(gam)%

c)

Gọi $n_X = a(mol) \Rightarrow n_{Al} = 1,5a(mol)$
Theo PTHH : 

$n_{H_2} = a + 1,5a.1,5 = 0,65 \Rightarrow a = 0,2(mol)$

Suy ra : $0,2.X + 0,2.1,5.27 = 12,9 \Rightarrow X = 24$

Vậy X là Magie