Question

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\(^{-1}\).

• A. 235.7 g
• B. 0.24 mg
• C. 236 mg
• D. 2.444 g

Hint:

Use the stoichiometry of the balanced chemical equation to relate the moles of the reactant and the product. At STP, 1 mole of a gas occupies 22.4 L. Convert the given volume of hydrogen gas to moles and then use the molar mass of magnesium to find the mass of magnesium required. Be careful with unit conversions (mL to L, g to mg).

Answer 1

The Correct Option is C

The balanced chemical equation for the reaction of magnesium with dilute HCl is: \[ \text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g) \] From the stoichiometry of the reaction, 1 mole of Mg produces 1 mole of H\( _2 \) gas. At STP (Standard Temperature and Pressure), 1 mole of any gas occupies a volume of 22.4 L or 22400 mL. Given volume of H\( _2 \) gas produced = 220 mL. Number of moles of H\( _2 \) produced = \( \frac{\text{Volume of H}_2 \text{ at STP}}{22400 \, \text{mL/mol}} \) \[ \text{Moles of H}_2 = \frac{220 \, \text{mL}}{22400 \, \text{mL/mol}} = \frac{22}{2240} \, \text{mol} = \frac{11}{1120} \, \text{mol} \] Since 1 mole of Mg produces 1 mole of H\( _2 \), the number of moles of Mg required is equal to the number of moles of H\( _2 \) produced. \[ \text{Moles of Mg used} = \frac{11}{1120} \, \text{mol} \] Given molar mass of Mg = 24 g mol\(^{-1}\). Mass of Mg required = Moles of Mg × Molar mass of Mg \[ \text{Mass of Mg} = \frac{11}{1120} \, \text{mol} \times 24 \, \text{g/mol} = \frac{264}{1120} \, \text{g} \] \[ \text{Mass of Mg} \approx 0.2357 \, \text{g} \] To convert grams to milligrams, multiply by 1000: \[ \text{Mass of Mg} \approx 0.2357 \times 1000 \, \text{mg} = 235.7 \, \text{mg} \] The closest option to 235.7 mg is 236 mg.