Question

Amount of magnesium (Mg) (in mg) required to liberate 224 mL of \( H_2 \) gas at STP, when reacted with HCl.

• A. 20
• B. 10
• C. 15
• D. 5

Hint:

To solve such problems, use the molar volume of a gas at STP (22.4 L) and stoichiometric relationships between reactants and products.

Answer 1

The Correct Option is A

The reaction between magnesium and hydrochloric acid is: \[ \text{Mg (s)} + 2 \, \text{HCl (aq)} \rightarrow \text{MgCl}_2 \, (aq) + \text{H}_2 \, (g) \] From the reaction, we see that 1 mole of magnesium produces 1 mole of hydrogen gas. At STP, 1 mole of gas occupies 22.4 L. Therefore, 224 mL (or 0.224 L) of \( H_2 \) gas corresponds to: \[ \text{moles of } H_2 = \frac{0.224 \, \text{L}}{22.4 \, \text{L/mol}} = 0.01 \, \text{mol} \] Since the molar ratio of magnesium to hydrogen gas is 1:1, 0.01 mol of hydrogen requires 0.01 mol of magnesium. The molar mass of magnesium (Mg) is 24 g/mol. Thus, the mass of magnesium required is: \[ \text{mass of Mg} = 0.01 \, \text{mol} \times 24 \, \text{g/mol} = 0.24 \, \text{g} = 240 \, \text{mg} \] Thus, the amount of magnesium required is 240 mg.