MHT CET · Chemistry · Redox Reactions
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 \(\mathrm{Mg}=24 \mathrm{~g} \mathrm{~mol}^{-1}\) )
- A \(12 \mathrm{~g}\)
- B \(4.8 \mathrm{~g}\)
- C \(6 \mathrm{~g}\)
- D \(2.4 \mathrm{~g}\)
Answer & Solution
Correct Answer
(B) \(4.8 \mathrm{~g}\)
Step-by-step Solution
Detailed explanation
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
\(=0.2 \mathrm{~mol}=0.2 \times 24=4.8 \mathrm{~g}\)
\(=\frac{\text { Volume of gas at STP }}{22.4 \mathrm{dm}^3 \mathrm{~mol}^{-1}}\)
\(\therefore \quad \mathrm{n}=\frac{4.48 \mathrm{dm}^3}{22.4 \mathrm{dm}^3 \mathrm{~mol}^{-1}}=0.2 \mathrm{~mol}\)
\(\mathrm{Mg}_{(\mathrm{s})}+2 \mathrm{HCl}_{(\mathrm{aq})} \longrightarrow \mathrm{MgCl}_2+\mathrm{H}_{2(\mathrm{~g})} \uparrow\)
\(1 \mathrm{~mol} \mathrm{Mg} \equiv 1 \mathrm{~mol} \mathrm{H}_2\) gas
\(\therefore \quad\) Mg required to liberate \(0.2 \mathrm{~mol} \mathrm{H}_2\) gas
\(=0.2 \mathrm{~mol}=0.2 \times 24=4.8 \mathrm{~g}\)
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