Mass of magnesium required to produce 220 mL of hydrogen gas at STP on reaction with excess of dil. HCl is (Given: Molar mass of Mg is 24 g mol$$^{-1}$$)

The balanced reaction of magnesium with dilute hydrochloric acid is
$$Mg + 2\,HCl \rightarrow MgCl_2 + H_2$$

From the equation, $$1$$ mole of $$Mg$$ produces $$1$$ mole of $$H_2$$.

Step 1 - Convert volume of $$H_2$$ to moles
At STP, $$1$$ mole of any ideal gas occupies $$22.4$$ L $$= 22\,400$$ mL.
Given volume of $$H_2$$ $$= 220$$ mL.

Number of moles of $$H_2$$ produced
$$n_{H_2}= \frac{220 \text{ mL}}{22\,400 \text{ mL mol}^{-1}}$$

$$n_{H_2}= 9.821\times10^{-3} \text{ mol}$$ (keep four significant figures).

Step 2 - Relate moles of $$H_2$$ to moles of $$Mg$$
The stoichiometry is $$1:1$$.
Therefore,
$$n_{Mg}= n_{H_2}= 9.821\times10^{-3} \text{ mol}$$.

Step 3 - Calculate mass of magnesium
Molar mass of $$Mg = 24 \text{ g mol}^{-1}$$.

Mass of $$Mg$$ required
$$m_{Mg}= n_{Mg}\times M_{Mg}= (9.821\times10^{-3})\times24$$

$$m_{Mg}= 2.357\times10^{-1} \text{ g}= 0.2357 \text{ g}$$.

Convert grams to milligrams:
$$0.2357 \text{ g}= 235.7 \text{ mg}\approx 236 \text{ mg}$$.

Hence, the mass of magnesium needed is approximately $$236$$ mg.
Correct option: Option C.