QJEEM22Ch1a8c: JEE Main 2022 Chemistry - Class 11 JEEM22Ch1a8c Solution: A 2.0 g sample containing MnO 2 is treated with HCl liberating…

First Step :

MnO₂ + 4HCl $\to$ MnCl₂ + Cl₂ + 2H₂O

Here 1 mol of MnO₂ produce 1 mol of Cl₂

$\therefore$ Mole ratio of ${n_{Mn{O_2}}}:{n_{C{l_2}}} = 1:1$

Second Step :

Cl₂ + 2KI $\to$ 2KCl + I₂

Here, 1 mol of Cl₂ produce 1 mol of I₂

Mole ratio of ${n_{C{l_2}}}:{n_{{I_2}}} = 1:1$

Third Step :

I₂ + 2Na₂S₂O₃ $\to$ 2NaI + Na₂S₂O₃

1 mol of I₂ react with 2 mol of Na₂S₂O₃

Mole ratio of ${n_{{I_2}}}:{n_{{N_2}{S_2}{O_3}}} = 1:2$

Given Na₂S₂O₃ is 60 mL of 0.1 M

$\therefore$ Number of moles of Na₂S₂O₃

= V (in L) $\times$ M (Molarity)

= ${{60} \over {1000}} \times 0.1$

= 0.006 mol

$\therefore$ Number of moles of I₂

$= {1 \over 2}(0.006)$

= 0.003

$\therefore$ Moles of MnO₂ = 0.003 (as mole ratio of MnO₂ and Cl₂ = 1 : 1)

Molar mass of MnO₂ = 55 + 32 = 87

$\therefore$ Mass of MnO₂ = 0.003 $\times$ 87 = 0.261 gm

Given MnO₂ = 2g

$\therefore$ % of MnO₂ = ${{0.261} \over 2} \times 100 = 13\%$