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

To ascertain the mass of magnesium (Mg) needed to yield 220 mL of hydrogen gas (H₂) at Standard Temperature and Pressure (STP) when reacting with excess dilute hydrochloric acid (HCl), we begin with the balanced chemical equation:

\(\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_{2} + \text{H}_2 \uparrow\)

1. The stoichiometry dictates that 1 mole of magnesium produces 1 mole of hydrogen gas.
2. At STP, 1 mole of an ideal gas occupies 22.4 L (equivalent to 22,400 mL).
3. Given the production of 220 mL of hydrogen, the moles of hydrogen gas are calculated as:

\(\text{Moles of H}_2 = \frac{220 \, \text{mL}}{22400 \, \text{mL/mol}} = 0.00982 \, \text{mol}\)

1. As the molar ratio of Mg to H₂ is 1:1, the moles of magnesium required are also 0.00982 mol.
2. With the molar mass of magnesium (Mg) being 24 g/mol, the required mass of magnesium is:

\(\text{Mass of Mg} = 0.00982 \, \text{mol} \times 24 \, \text{g/mol} = 0.23568 \, \text{g}\)

1. Conversion to milligrams (1 g = 1000 mg) yields:

\(0.23568 \, \text{g} = 235.68 \, \text{mg} \approx 236 \, \text{mg}\)

Consequently, the mass of magnesium required is approximately 236 mg.

Conclusion: The calculated mass is 236 mg.