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