$\rm n_{HCl}=0,24.1=0,24(mol)$

$\rm a.PTHH: A_2O_3+6HCl \to 2ACl_3+3H_2O$

$\rm b.n_{A_2O_3}=\dfrac{1}{6}n_{HCl}=0,04(mol)$

$\rm M_{A_2O_3}=\dfrac{6,4}{0,04}=160(amu)$

$\rm \to 2A+16.3=160$

`=>`$\rm A=56(amu)$

vậy $\rm A:iron (Fe)$

vậy CTHH của oxide: $\rm Fe_2O_3$

a)PTHH

$\rm A_2O_3 + 6HCl \to 2ACl_3 + 3H_2O$

b)Đổi $\rm 240ml = 0,24L$

$\rm \to n_{HCl} = 0,24 \times 1 = 0,24$ mol

Theo PTHH

$\rm n_{A_2O_3} = \dfrac{1}{6}n_{HCl} = \dfrac{1}{6}\times 0,24 = 0,04$ mol

$\rm \to M_{A_2O_3} = 2M_A + 16 \times 3 = \dfrac{6,4}{0,04}$

$\rm \to M_A = 56(Fe)$

Vậy oxide là $\rm Fe_2O_3$