\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{1,68}{22,4}=0,075mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,15 0,075 0,075 ( mol )
\(V_{HCl}=\dfrac{n_{HCl}}{C_{M\left(HCl\right)}}=\dfrac{0,15}{0,5}=0,3l=300ml\)
\(m_{FeCl_2}=n_{FeCl_2}.M_{FeCl_2}=0,075.127=9,525g\)