n\(_{HCl}\) = 5 *0,5 = 2,5 (mol)
⇒ m\(_{HCl}\) = 2,5 *36,5 = 91,25 (gam)
⇒ m\(_{ddHCl}\) = \(\dfrac{91,25\cdot100\%}{36\%}\) \(\sim\) 253,5 (gam)
\(\Rightarrow\) V\(_{HCl}\) = \(\dfrac{m_{ddHCl}}{D_{HCl}}\)= \(\dfrac{253,5}{1,19}\) \(\sim\) 230(milit)