The ratio of Fe3+ and Fe2+ ions in Fe0.9S1.0 is

Formula: Fe0.9O1.0 ≅ Fe90O100Let,Number of moles of Fe+2 ions =X Then, number of moles of Fe+3 = [ 90-X ]we now that,Sum of the charges of all the atoms in a compound = 0X+2+90-X+3+100-2=0 2X-3X+270-200=0 70-X=0 XMoles of Fe+2 ions=70 Moles of Fe+3 ions=90-X=90-70=20 Fe+3Fe+2=2070=0.28