$$a)~PTHH:~Fe~+~2HCl-->FeCl_2~+~H_2\uparrow$$
$$b)~n_{H_2}=\frac{V}{22,4}=$$ $$\frac{3,36}{22,4}=0,15~(~mol~)$$
     $$n_{H_2}=n_{Fe}=0,15~(~mol~)$$
$$-->~m_{Fe}=n~.~M=0,15~.~56=8,4~(~g~)$$
$$c)~$$$$C_{M~dung~dịch~HCl}=\frac{n}{V}=$$ $$\frac{0,15~.~2}{0,05}=6~M$$ 

a)
n_{H_2}={3,36}/{22,4}=0,15(mol)
Fe+2HCl->FeCl_2+H_2
0,15←0,3←0,15←0,15(mol)
m_{Fe}=0,15.56=8,4(g)
c)
C_{M\ HCl}={0,3}/{0,05}=6M