Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
According to reaction,
$\mathrm{Mg}_{(\mathrm{s})}+2 \mathrm{HCl}_{(\mathrm{aq})} \longrightarrow \mathrm{MgCl}_{2(\mathrm{aq})}+\mathrm{H}_{2(\mathrm{~g})} \uparrow$
Calculate the mass of $\mathrm{Mg}$ required to liberate
$4.48 \mathrm{dm}^3 \mathrm{H}_2$ at STP.
$\left(\right.$ Molar mass of $\mathrm{Mg}=24 \mathrm{~g} \mathrm{~mol}^{-1}$ )
ChemistryRedox ReactionsMHT CETMHT CET 2023 (13 May Shift 2)
Options:
  • A $12 \mathrm{~g}$
  • B $4.8 \mathrm{~g}$
  • C $6 \mathrm{~g}$
  • D $2.4 \mathrm{~g}$
Solution:
1680 Upvotes Verified Answer
The correct answer is: $4.8 \mathrm{~g}$
Number of moles of a gas (n)
$=\frac{\text { Volume of gas at STP }}{22.4 \mathrm{dm}^3 \mathrm{~mol}^{-1}}$
$\therefore \quad \mathrm{n}=\frac{4.48 \mathrm{dm}^3}{22.4 \mathrm{dm}^3 \mathrm{~mol}^{-1}}=0.2 \mathrm{~mol}$
$\mathrm{Mg}_{(\mathrm{s})}+2 \mathrm{HCl}_{(\mathrm{aq})} \longrightarrow \mathrm{MgCl}_2+\mathrm{H}_{2(\mathrm{~g})} \uparrow$
$1 \mathrm{~mol} \mathrm{Mg} \equiv 1 \mathrm{~mol} \mathrm{H}_2$ gas
$\therefore \quad$ Mg required to liberate $0.2 \mathrm{~mol} \mathrm{H}_2$ gas
$=0.2 \mathrm{~mol}=0.2 \times 24=4.8 \mathrm{~g}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.