1.
Gọi a, b là mol Ag, Cu.
$\Rightarrow 108a+64b=1,12$ (1)
$n_{SO_2}= \frac{n_{Ag}+2n_{Cu}}{2}= 0,5a+b$ (bảo toàn e)
$n_{BaSO_4}=\frac{1,864}{233}=0,008 mol$
$SO_2+Cl_2+2H_2O\to 2HCl+H_2SO_4$
$BaCl_2+H_2SO_4\to BaSO_4+2HCl$
$\Rightarrow 0,5a+b=0,008$ (2)
(1)(2) $\Rightarrow a=0,008; b=0,004$
$\%m_{Ag}=\frac{0,008.108.100}{1,12}=77,14\%$
$\%m_{Cu}=22,86\%$
b,
$n_{NaOH}=0,028.0,5=0,014 mol$
$\frac{n_{NaOH}}{n_{SO_2}}= 1,75$
$\Rightarrow$ Tạo 2 muối: $NaHSO_3$ (a mol), $Na_2SO_3$ (b mol)
Bảo toàn Na: $a+2b=0,014$ (1)
Bảo toàn S: $a+b=0,008$ (2)
(1)(2) $\Rightarrow a=0,002; b=0,006$
$m_{NaHSO_3}=0,002.104=0,208g$
$m_{Na_2SO_3}=0,006.126=0,756g$