Practical Organic Chemistry

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=‌

0.261

2

×100=13%