Mass of magnesium required to produce 220 mL of hydrogen gas at STP on reaction with excess of dil. HCl is: (Given:…

🧠 One Mg makes one $\ce{H2}$ The reaction is $\ce{Mg + 2HCl -> MgCl2 + H2}$. So $1$ mole of Mg gives $1$ mole of $\ce{H2}$. Find moles of gas, then mass of Mg.

🗺️ Step by step Moles of $\ce{H2}$ at STP: $n = \frac{220}{22400} \approx 9.82 \times 10^{-3}\,\text{mol}.$

Moles of Mg needed $= 9.82 \times 10^{-3}\,\text{mol}$ (same as $\ce{H2}$).

Mass of Mg: $0.00982 \times 24 \approx 0.2357\,\text{g} = 235.7\,\text{mg}.$

The closest option is $236\,\text{mg}$.

⚡ Fast check $220/22400 \approx 1/100$ mol. $1/100 \times 24 = 0.24\,\text{g} = 240\,\text{mg}$. So roughly $236\,\text{mg}$ is the right scale.

$\boxed{\text{Answer: (c)}}$