In the reaction 2Al(s) + 6HCl(aq) → 2Al^3+(aq) + 6Cl^-(aq) + 3H_2(g) , which statement is correct? - Chemistry MCQ AIEEE 2007 2026 | ExamTest.live

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In the reaction 2 Al (s) + 6 HCl (a q) \to 2 Al^{3 +} (a q) + 6 Cl^{-} (a q) + 3 H_{2} (g), which statement is correct?

Solution & Step-by-step Explanation

From the balanced equation: 6 moles of HCl produce 3 moles of H_{2} gas.Therefore, 1 mole of HCl produces 0.5 mole of H_{2} gas.At STP, 1 mole of an ideal gas occupies 22.4 L .Volume of H_{2} produced per mole of HCl = 0.5 \times 22.4 = 11.2 L .Check other options:Stoichiometry for Al : 2 moles Al \to 3 moles H_{2} . So 1 mole Al \to 1.5 moles H_{2} = 33.6 L at STP (not regardless of T, P).Option A refers to volume of solution vs volume of gas, which depends on molarity.

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In the reaction 2 Al (s) + 6 HCl (a q) \to 2 Al^{3 +} (a q) + 6 Cl^{-} (a q) + 3 H_{2} (g), which statement is correct?

6 L HCl (a q) is consumed for every 3 L H_{2} (g) produced

33.6 L H_{2} (g) is produced for every mole of Al that reacts (at any T, P)

67.2 L H_{2} (g) at STP is produced for every mole Al that reacts

11.2 L H_{2} (g) at STP is produced for every mole HCl (a q) consumed

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