5g of pyrolusite (impure MnO2) were heated with conc. HCl and Cl2evolved was passed through excess of KI solution. The iodine liberated required 40 ml of N/10 hypo solution. Find the % of MnO2in the pyrolusite

The iodine will react with 40 ml of 1/10 N of hypo solution I2 +2Na2S2O3-----> 2NaI+ Na2S4O6 . So n factor of sodium thiosulfate is 2. Molarity =normality/n factor =1/5 . So moles of sodium thiosulfate in 40 ml = 0.008 moles . According to equation I2 reacting with sodium thiosulphate =0.004 moles. So iodine produced by chlorine reacting with potassium iodide =0.004 mol now Cl2+ 2KI------>I2 + 2KCl as moles of chlorine used are same as moles of iodine produced moles of chlorine liberated by HCl =0.004 moles . Now MnO2 + 4HCl----->2Cl2 + Mn + 2H2O so moles of manganese oxide =1/2 moles of chlorine =0.002 moles now mass of pure MnO2 = 0.002×(55+32)= 87×0.002 = 0.174 now % of pure MnO2 in pyrolusite ore = 0.174/5 ×100 = 3.48%