\(n_{HCl}=0,2\left(mol\right);n_{H_2SO_4}=0,3\left(mol\right);n_{H_2}=0,4\left(mol\right)\\ BTNT.H\Rightarrow2n_{H_2}=n_{HCl\left(pứ\right)}+n_{H_2SO_4\left(pứ\right)}\\ =0,8=n_{HCl}+n_{H_2SO_4}\\ \Rightarrow HCl;H_2SO_4\text{ pứ hết.}\\ \Rightarrow m_{Muối}=m_{KL}+m_{Cl^-}+m_{SO_4^{2-}}=43,7\left(g\right) \)