a,
Đặt $x$, $y$ là số mol $Fe$, $Cu$
$\to 56x+64y=6$ $(1)$
$n_{H_2SO_4}=\dfrac{25.98\%}{98}=0,25(mol)$
$2Fe+6H_2SO_4\to Fe_2(SO_4)_3+3SO_2+6H_2O$
$Cu+2H_2SO_4\to CuSO_4+SO_2+2H_2O$
$\to 3x+2y=0,25$ $(2)$
$(1)(2)\to x=y=0,05$
$m_{Fe}=56x=2,8g$
$m_{Cu}=64y=3,2g$
b,
$n_{SO_2}=1,5x+y=0,125(mol)$
$V_{dd}=100cm^3=0,1dm^3=0,1l$
$\to n_{NaOH}=0,1.3=0,3(mol)$
$SO_2+2NaOH\to Na_2SO_3+H_2O$
$\to NaOH$ dư
$n_{Na_2SO_3}=n_{SO_2}=0,125(mol)$
$\to C_{M_{Na_2SO_3}}=\dfrac{0,125}{0,1}=1,25$
$n_{NaOH\text{dư}}=0,3-0,125.2=0,05(mol)$
$\to C_{M_{NaOH}}=\dfrac{0,05}{0,1}=0,5M$
