Question

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 :

• A. 2 mol of HCl (g)
• B. 0 .5 mol of HCl (g)
• C. 1.5 mol of HCl (g)
• D. 1 mol of HCl (g)

Answer 1

The Correct Option is D

1. At Standard Temperature and Pressure (S.T.P.), 1 mole of any gas occupies 22.4 litres. This is a key point for solving this problem using the concept of mole and volume relation of gases at S.T.P.
2. Determine the moles of each gas:
   • H_2: Given 22.4 litres
     • Moles of H_2 = \frac{22.4 \, \text{litres}}{22.4 \, \text{litres/mole}} = 1 \, \text{mole}
   • Cl_2: Given 11.2 litres
     • Moles of Cl_2 = \frac{11.2 \, \text{litres}}{22.4 \, \text{litres/mole}} = 0.5 \, \text{mole}
3. The balanced chemical reaction for the formation of hydrogen chloride is:

   H_2(g) + Cl_2(g) \rightarrow 2HCl(g)

   • According to the reaction, 1 mole of H_2 reacts with 1 mole of Cl_2 to produce 2 moles of HCl.
4. Identify the limiting reactant:
   • Since only 0.5 moles of Cl_2 are available while we have 1 mole of H_2, Cl_2 is the limiting reactant.
5. Calculate the moles of HCl formed:
   • According to the stoichiometry of the balanced equation, 0.5 moles of Cl_2 will react with 0.5 moles of H_2 to produce 1 mole of HCl.
6. Conclusion: The moles of HCl formed is equal to 1 mole.