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 $24 \mathrm{~g} \mathrm{~mol}^{-1}$.

<p>We are tasked with calculating the mass of magnesium required to produce <strong>220 mL of hydrogen gas at STP</strong> with excess dilute HCl. </p> <p><strong>Step 1: Write the balanced reaction</strong> </p> <p>$ \text{Mg} + 2 \text{HCl} \;\;\rightarrow\;\; \text{MgCl}_2 + \text{H}_2 $</p> <ul> <li>1 mole Mg gives 1 mole H₂.</li> </ul> <p><strong>Step 2: Molar volume at STP</strong> </p> <p>At STP, 1 mole of an ideal gas occupies <strong>22.4 L = 22,400 mL</strong>.</p> <p><strong>Step 3: Moles of H₂ produced</strong> </p> <p>$ n(\text{H}_2) = \frac{\text{Volume of H₂}}{\text{Molar Volume}} = \frac{220}{22400} = 0.00982 \,\text{mol} $</p> <p><strong>Step 4: Moles of Mg required</strong> </p> <p>From stoichiometry, 1 mole Mg produces 1 mole H₂, so: </p> <p>$ n(\text{Mg}) = n(\text{H}_2) = 0.00982 \,\text{mol} $</p> <p><strong>Step 5: Mass of Mg required</strong> </p> <p>$ m(\text{Mg}) = n \times M(\text{Mg}) = 0.00982 \times 24 = 0.236 \,\text{g} $</p> <p>$ = 236 \,\text{mg} $</p> <h3>✅ Final Answer:</h3> <p><strong>Option D: 236 mg</strong></p>