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When 22.4 litres of H2(g) is mixed with 11.2 litres of Cl2(g), each at S.T.P., the moles of HCl(g) formed is equal to:

1. 1 mole of HCl(g)
2. 2 moles of HCl(g)
3. 0.5 moles of HCl(g)
4. 1.5 moles of HCl(g)

Correct answer: 1 mole of HCl(g)

Solution

At STP, 1 mole of any gas occupies 22.4 L. The reaction H2 + Cl2 → 2HCl shows a 1:1 molar ratio between H2 and Cl2. Since 11.2 L of Cl2 (0.5 moles) is the limiting reagent, it reacts with 0.5 moles of H2 to form 1 mole of HCl.

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