a,

$n_{HCl}=0,075.2=0,15 mol$

$Fe_2O_3+6HCl\to 2FeCl_3+3H_2O$

$\Rightarrow n_{Fe_2O_3}=0,025 mol$

$m_M=8,8-0,025.160=4,8g$

M là kim loại khử yếu nên A là $SO_2$

$n_{SO_2}=\dfrac{1,68}{22,4}=0,075 mol$

$Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O$

$M+2H_2SO_4\to MSO_4+SO_2+H_2O$

$\Rightarrow n_M=0,075 mol$

$M_M=\dfrac{4,8}{0,075}=64(Cu)$

b,

$\%m_{Cu}=\dfrac{4,8.100}{8,8}=54,5\%$

$\%m_{Fe_2O_3}=45,5\%$

c,

$n_{NaOH}=\dfrac{16,8.1,25.20\%}{40}=0,105 mol$

$\dfrac{n_{NaOH}}{n_{SO_2}}=1,4$

$\to$ tạo muối $NaHSO_3$ (x mol), $Na_2SO_3$ (y mol)

Bảo toàn Na: $x+2y=0,105$   (1)

Bảo toàn S: $x+y=0,075$       (2)

(1)(2)$\Rightarrow x=0,045; y=0,03$

$m_{NaHSO_3}=0,045.104=4,68g$

$m_{Na_2SO_3}=0,03.126=3,78g$