Question

MnO2 is known to oxidise HCI to Cl2 and itself gets reduced to MnCl2. 50 mL of 0.4 M HCl is treated with excess of MnO2. If is the volume of Cl2 gas evolved is x mL at STP (1 atm and 273 K) then the value of $\frac{x}{10}$ (nearest integer)

Answer 1

11

Answer 2

Solution:

1. Balanced Equation:

   $\text{MnO}_2 + 4 \text{HCl} \rightarrow \text{MnCl}_2 + \text{Cl}_2 + 2 \text{H}_2\text{O}$

2. Moles of HCl:

   Volume = 50 mL = 0.05 L
   Molarity = 0.4 M
   Moles = $0.05 \times 0.4 = 0.02$ moles

3. Moles of Cl₂ Produced:

   From the equation, 4 moles HCl give 1 mole Cl₂
   Moles of Cl₂ = $\frac{0.02}{4} = 0.005$ moles

4. Volume of Cl₂ at STP:

   At STP, 1 mole gas = 22.4 L = 22400 mL
   Volume = $0.005 \times 22400 = 112\, \text{mL}$
   Thus, $x = 112\, \text{mL}$.

5. Final Calculation:

   $\frac{x}{10} = \frac{112}{10} = 11.2$
   Rounded to nearest integer: 11

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Explanation (minimal):
HCl moles = 0.02; reaction stoichiometry gives Cl₂ moles = 0.005; volume of Cl₂ at STP = 112 mL; hence, x/10 = 11.2 ≈ 11.