\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)
a)\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 0,1 0,1
\(m_{Fe}=0,1\cdot56=5,6\left(g\right)\)
b)\(\Rightarrow\%m_{Fe}=\dfrac{5,6}{12}\cdot100\%=46,67\%\) \(\Rightarrow\%m_{Cu}=100\%-46,67\%=53,33\%\)
c)\(n_{NaOH}=0,1\cdot1=0,1mol\)
\(2NaOH+FeCl_2\rightarrow Fe\left(OH\right)_2+2NaCl\)
0,1 0,1 0,1
\(m_{Fe\left(OH\right)_2}=0,1\cdot90=9\left(g\right)\)