1.
\(\%C=\dfrac{12}{44}.100\simeq22,73\%\)
2.
\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ V=\dfrac{n}{C_M}=\dfrac{0,1}{2}=0,05\left(M\right)\)
\(C_{M\left(FeCl_2\right)}=\dfrac{n}{V}=\dfrac{0,05}{0,05}=1\left(M\right)\)