Lời giải :

.

$n_{HCl} = V.C_M = 0,15.2,5 = 0,375(mol)$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$

Theo PTHH : 

$n_{Al} = \dfrac{2}{6}n_{HCl} = 0,125(mol)$
$\Rightarrow m = 0,125.27 = 3,375(gam)$

$n_{HCl} = 0,3.1 = 0,3(mol)$
$NaOH + HCl \to NaCl + H_2O$

Theo PTHH : 

$n_{NaOH} = n_{HCl} = 0,3(mol)$
$V_{dd\ NaOH} = \dfrac{n}{C_M} = \dfrac{0,3}{0,5} = 0,6(lít)$

$n_{Ca} = \dfrac{12}{40} = 0,3(mol)$
$Ca + 2HCl \to CaCl_2 + H_2$

Theo PTHH : 

$n_{CaCl_2} = n_{Ca} = 0,3(mol)$
$m_{CaCl_2} = 0,3.111 = 33,3(gam)$

Đáp án : $Ba(OH)_2, Ba(NO_3)_2$

$Ba(OH)_2+ + H_2SO_4 \to BaSO_4 + 2H_2O$
$Ba(NO_3)_2 + H_2SO_4 \to BaSO_4 + 2HNO_3$