First Step : MnO2+4‌HCl→MnCl2+Cl2+2H2O Here 1‌mol of MnO2 produce 1‌mol of Cl2 ∴ Mole ratio of nMnO2:nCl2=1:1 Second Step : Cl2+2‌Kl→2‌KCl+I2 Here, 1‌mol of Cl2 produce 1‌mol of I2 Mole ratio of nCl2:nI2=1:1 Third Step : I2+2Na2S2O3→2‌Nal+Na2S2O3 1‌mol of I2 react with 2‌mol of Na2S2O3 Mole ratio of nI2:nN2S2O3=1:2 Given Na2S2O3 is 60‌mL of 0.1M ∴ Number of moles of Na2S2O3 =V( in L)×M (Molarity) =‌
60
1000
×0.1 =0.006‌mol ∴ Number of moles of I2 =‌
1
2
(0.006) =0.003 ∴ Moles of MnO2=0.003 (as mole ratio of MnO2 and Cl2=1:1 ) Molar mass of MnO2=55+32=87 ∴‌ Mass of ‌MnO2=0.003×87=0.261‌gm Given MnO2=2g ∴%‌ of ‌MnO2=‌