$a.PTHH : \\Fe+2HCl\to FeCl_2+H_2↑ \\n_{H_2}=\dfrac{4,48}{22,4}=0,2mol \\Theo\ pt : \\n_{Fe}=n_{H_2}=0,2mol \\⇒m_{Fe}=0,2.56=11,2g \\m_{Cu}=17,6-11,2=6,4g \\b.m_{H_2}=0,2.2=0,4g \\m_{dd\ spu}=17,6+100-0,4-6,4=110,8g \\Theo\ pt : \\n_{FeCl_2}=n_{H_2}=0,2mol \\⇒m_{FeCl_2}=0,2.127=25,4g \\⇒C\%_{FeCl_2}=\dfrac{25,4}{110,8}.100\%=22,92\%$