Answer:
[tex]\boxed{\text{199 mL}}[/tex]
Explanation:
1. Calculate the moles of HCl
[tex]n = \text{25.4 g} \times \dfrac{\text{1 mol}}{\text{36.46 g}} = \text{0.6967 mol}[/tex]
2. Calculate the volume of dilute HCl
[tex]\begin{array}{rcl}\text{Molar concentration}& =& \dfrac{\text{moles}}{\text{litres}}\\\\3.50 & = & \dfrac{0.6967}{V}\\\\3.50V & = & 0.6967\\\\V & = & \dfrac{0.6967}{3.50}\\\\V & = & \text{0.199 L} =\textbf{199 mL}\end{array}\\\text{The volume of muriatic acid needed is }\boxed{\textbf{199 mL}}[/tex]
