$$2HCl + Mg \rightarrow MgCl_2 + H_2$$

From the equation, we can see that 2 moles of HCl are required to produce 1 mole of hydrogen gas. Therefore, the number of moles of hydrogen gas produced from 250.0 milliliters of 3.0 M HCl can be calculated as follows:

$$Molarity = \frac{Moles \ of \ Solute}{Liters \ of \ Solution}$$

$$Moles \ of \ HCl = Molarity \times Liters \ of \ Solution$$

$$Moles \ of \ HCl = 3.0 M \times 0.250 L = 0.750 \ moles \ HCl$$

$$Moles \ of \ H_2 = \frac{0.750 \ moles \ HCl}{2} = 0.375 \ moles \ H_2$$

Therefore, 0.375 moles of hydrogen gas will be produced from 250.0 milliliters of 3.0 M HCl in an excess Mg.