\(n_{HCl}=n_{H2SO4}=0,5.0,2=0,1(mol)\\ m_{HCl}=0,1.36,5=3,65g\\ m_{H_2SO_4}=0,1.98=9,8g\\ n_H=0,1+0,1.2=0,3(mol)\\ →n_{H_2O}=\frac{0,3}{2}=0,15(mol)\\ m_{H_2O}=0,15.18=2,7g\\ BTKL:\\ m_{muối}=12+3,65+9,8-2,7=22,75g\\ →B\)
\(n_{HCl}=n_{H2SO4}=0,5.0,2=0,1(mol)\\ m_{HCl}=0,1.36,5=3,65(g)\\ m_{H2SO4}=0,1.98=9,8(g)\\ BTKL:\\ m_{muối}=12+3,65+9,8=25,45g\\ \to A\)