\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ 2K+2H_2O\rightarrow2KOH+H_2\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ KOH+HCl\rightarrow KCl+H_2O\\ NaOH+HCl\rightarrow NaCl+H_2O\\ n_{HCl}=n_{KOH}+n_{NaOH}=2.n_{H_2}=2.0,15=0,3\left(mol\right)\\ m_{HCl}=0,3.36,5=10,95\left(g\right)\)