Given data:

Volume of H₂: 220 mL

Molar mass of Mg: $24 \, \mathrm{g/mol}$

At STP, 1 mole of a gas occupies 22.4 liters (or 22,400 mL).

First, determine the moles of hydrogen gas ($\mathrm{H}_2$) produced:

$ \text{Moles of } \mathrm{H}_2 = \frac{220 \times 10^{-3}}{22.4} $

Because the molar ratio of $\mathrm{Mg}$ to $\mathrm{H}_2$ in the reaction is 1:1, the moles of $\mathrm{Mg}$ used are equal to the moles of $\mathrm{H}_2$:

Next, calculate the mass of magnesium:

$ \text{Mass of Mg used} = \frac{220 \times 10^{-3}}{22.4} \times 24 $

Simplifying further:

$ = 235.7 \times 10^{-3} \, \mathrm{g} $

Convert the mass from grams to milligrams:

$ = 235.7 \, \mathrm{mg} $

Thus, the mass of magnesium required is 235.7 mg.