In the reaction. $2 \mathrm{Al}_{(\mathrm{s})}+6 \mathrm{HCl}_{(\mathrm{s})} \longrightarrow 2 \mathrm{Al}^{3+}_{(\mathrm{aq})}+6 \mathrm{Cl}^{-}_{(\mathrm{aq})}+3 \mathrm{H}_{2(\mathrm{~g})}$,

(D) $11.2 \mathrm{H}_{2(\mathrm{~g})}$ at STP is produced for every mole $\mathrm{HCl}_{(\mathrm{aq})}$ consumed

(A) $6 \mathrm{~L} \mathrm{HCl}_{(\mathrm{aq})}$ is consumed for every $3 \mathrm{~L} \mathrm{H}_{2(\mathrm{~g})}$ produced

(B) $33.6 \mathrm{~L} \mathrm{H}_{2(\mathrm{~g})}$ is produced regardless of temperature and pressure for every mole Al that reacts

(C) $67.2 \mathrm{~L} \mathrm{H}_{2(\mathrm{~g})}$ at STP is produced for every mole AI that reacts

JEE Main · 2007 · 29 Apr 2007 · Offline · Some Basic Concepts of Chemistry

In the reaction.

2 Al_{(s)} + 6 HCl_{(s)} ⟶ 2 Al_{(aq)}^{3 +} + 6 Cl_{(aq)}^{-} + 3 H_{2 (g)}

(A) $6 L HCl_{(aq)}$ is consumed for every $3 L H_{2 (g)}$ produced
(B) $33.6 L H_{2 (g)}$ is produced regardless of temperature and pressure for every mole Al that reacts
(C) $67.2 L H_{2 (g)}$ at STP is produced for every mole AI that reacts
(D) $11.2 H_{2 (g)}$ at STP is produced for every mole $HCl_{(aq)}$ consumed✓ Correct

Answer: (D)

The equation

2 Al + 6 HCl \to 2 A l^{3 +} + 6 C l^{-} + 3 H_{2}

shows six moles of acid producing three moles of hydrogen, a

2 : 1

ratio. One mole of hydrochloric acid therefore yields half a mole of hydrogen gas. At STP that half mole occupies

22.4 \times 0.5 = 11.2

litre for every mole of acid consumed.

JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option