\(Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\\ n_{Fe}=n_{H_2}=0,2\left(mol\right)\\ \Rightarrow m_B\left(m_{Cu}\right)=m_{Fe}=0,2.56=11,2\left(g\right)\\ n_{Cu}=\dfrac{11,2}{64}=0,175\left(mol\right)\\ PTHH:2Fe+6H_2SO_{4\left(đ\right)}\rightarrow\left(t^o\right)Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\\ Cu+2H_2SO_{4\left(đ\right)}\rightarrow\left(t^o\right)CuSO_4+SO_2+2H_2O\\ n_{SO_2}=\dfrac{3}{2}n_{Fe}+n_{Cu}=\dfrac{3}{2}.0,2+0,175=0,475\left(mol\right)\\ 5SO_2+2KMnO_4+2H_2O\rightarrow2MnSO_4+K_2SO_4+2H_2SO_4\\ n_{KMnO_4}=\dfrac{2}{5}.0,475=0,095\left(mol\right)\\ V_{ddKMnO_4}=\dfrac{0,095}{0,5}=0,19\left(M\right)\)
