`a)`

`Fe+2HCl->FeCl_2+H_2`

`Cu+4HNO_{3\ đ}->Cu(NO_3)_2+2NO_2+2H_2O`

`b)`

Theo PT:

`n_{Fe}=n_{H_2}={3,36}/{22,4}=0,15(mol)`

`n_{Cu}=1/2n_{NO_2}=1/{2}.{6,72}/{22,4}=0,15(mol)`

`->\%m_{Cu}={0,15.64}/{0,15.64+0,15.56}.100\%\approx 53,33\%`

`->\%m_{Fe}=100-53,33=46,67\%`

$Chỉ~Fe~tác~dụng~vs~HCl:$

$~Fe~+~2HCl-->FeCl_2~+~H_2\uparrow.$

$Chỉ~Cu~tác~dụng~với~HNO_3:$

$~Cu~+~4HNO_3-->Cu(NO_3)_2~+~2H_2O~+~2NO_2\uparrow$

$n_{H_2}=\frac{V}{22,4}=$ $\frac{3,36}{22,4}=0,15~(~mol~)$

    $-->m_{Fe}=n~.~M=0,15~.~56=8,4~(~g~)$

.

$n_{NO_2}=\frac{V}{22,4}=$ $\frac{6,72}{22,4}=0,3~(~mol~)$

    $-->m_{Cu}=n~.~M=0,15~.~64=9,6~(~g~)$

.

$m_{Kim~loại}=9,6~+~8,4=18~(~g~)$

$C$%$_{Fe}=\frac{8,4~.~100}{18}=46,7$%  

$C$%$_{Cu}=\frac{9,6~.~100}{18}=53,3$%