a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\) (1)
b) Ta có: \(n_{H_2}=\frac{2,24}{22,4}=0,1\left(mol\right)=n_{Fe}\)
\(\Rightarrow m_{Fe}=0,1\cdot56=5,6\left(g\right)\) \(\Rightarrow m_{Cu}=6,4\left(g\right)\)
c) PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\) (2)
Theo PTHH (1): \(n_{HCl}=2n_{Fe}=0,2mol\)
\(\Rightarrow n_{NaOH}=0,2mol\) \(\Rightarrow m_{NaOH}=0,2\cdot40=8\left(g\right)\) \(\Rightarrow m_{ddNaOH}=\frac{8}{20\%}=40\left(g\right)\)