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\(^{-1}\).
Show Hint
- 235.7 g
- 0.24 mg
- 236 mg
- 2.444 g
The Correct Option is C
Approach Solution - 1
To determine the mass of magnesium (Mg) required to produce 220 mL of hydrogen gas (H2) at Standard Temperature and Pressure (STP) when reacted with excess dilute hydrochloric acid (HCl), we start by considering the balanced chemical reaction:
\(\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_{2} + \text{H}_2 \uparrow\)
- The balanced equation shows that 1 mole of magnesium produces 1 mole of hydrogen gas.
- At STP, 1 mole of any ideal gas occupies 22.4 L (or 22,400 mL).
- Given that 220 mL of hydrogen is produced, we calculate the moles of hydrogen gas:
\(\text{Moles of H}_2 = \frac{220 \, \text{mL}}{22400 \, \text{mL/mol}} = 0.00982 \, \text{mol}\)
- Since 1 mole of Mg produces 1 mole of H2, the moles of magnesium required is also 0.00982 mol.
- The molar mass of magnesium (Mg) is given as 24 g/mol. Thus, the mass of magnesium required is:
\(\text{Mass of Mg} = 0.00982 \, \text{mol} \times 24 \, \text{g/mol} = 0.23568 \, \text{g}\)
- Converting grams to milligrams (1 g = 1000 mg):
\(0.23568 \, \text{g} = 235.68 \, \text{mg} \approx 236 \, \text{mg}\)
Thus, the mass of magnesium required is approximately 236 mg.
Conclusion: The correct answer is 236 mg.
Approach Solution -2
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