Question

The mass of magnesium required to produce 220 mL of hydrogen gas at STP on reaction with excess of dilute HCl is: (Given molar mass of Mg = 24 g/mol)

• A. 0.44 g
• B. 0.22 g
• C. 0.88 g
• D. 1.32 g

Hint:

In gas-related stoichiometric problems, always remember to use the molar volume of gases at STP (22.4 L or 22400 mL) to convert between volume and moles.

Answer 1

The Correct Option is A

The balanced equation for 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 equation, 1 mole of Mg produces 1 mole of H\(_2\) gas. We are given that the volume of hydrogen gas at STP is 220 mL, and we need to convert it to moles. 1. At STP, 1 mole of any ideal gas occupies 22.4 L (or 22400 mL). The number of moles of hydrogen gas is: \[ n_{\text{H}_2} = \frac{220}{22400} = 0.00982 \, \text{mol} \] 2. According to the balanced equation, 1 mole of Mg produces 1 mole of H\(_2\). Therefore, the moles of Mg required will be the same as the moles of H\(_2\), which is 0.00982 mol. 3. Now, we calculate the mass of magnesium required using the molar mass of Mg: \[ \text{Mass of Mg} = n_{\text{Mg}} \times \text{Molar mass of Mg} = 0.00982 \times 24 = 0.44 \, \text{g} \] Thus, the mass of magnesium required is 0.44 g.