How much concentrated solution would it take to prepare 2.90 L of 0.420 M HCl upon dilution with water

the quantity required can go from 117 ml (for maximum concentration) up to 2900 ml (if the concentrated solution has molarity = 0.420 M)

Explanation:

the amount of water required to dilute a solution V₁ liters of Molarity M₁ to V₂ liters of M₂

moles of hydrochloric acid = M₁ * V₁ = M₂ * V₂

V₁ = V₂ * M₂/M₁

where

M₂ = 0.420 M

V₂ = 2.90 L

Since the hydrochloric acid can be concentrated up to 38% p/V (higher concentrations are possible but the evaporation rate is so high that handling and storage require extra precautions, like cooling and pressurisation)

maximum M₁ = 38% p/V = 38 gr / 0.1 L / 36.5 gr/mol = 10.41 M

then

min V₁ = V₂ * M₂ / max M₁ = 2.90 L * 0.420 M / 10.41 M = 0.117 L = 117 ml

then the quantity required can go from 117 ml up to 2900 ml (if M₁ = M₂)