Does anyone know how to do this?
Metal ions Q2+ in acidified aqueous solution can be oxidised by aqueous potassiumdichromate(VI).In a titration, an acidified 25.0 cm3 sample of a 0.140 mol dm–3 solution of Q2+(aq)required 29.2 cm3 of a 0.040 mol dm–3 solution of potassium dichromate(VI) forcomplete reaction.Determine the oxidation state of the metal Q after reaction with the potassiumdichromate(VI)
This is the solution, it loses me at multiplying it by 6
moles of dichromate = (29.2/1000)×0.04 = 0.001168 or 0.00117 (1) moles of Q2+ = (25/1000)×0.140 = 0.0035(0) (1) each mole of dichromate needs 6 electrons or half equation with 6 e- (1) moles of electrons = 6×0.001168 = 0.007008 or moles Q2+:moles (1)dichromate = 3:1Moles of electrons per mole of Q = 0.007008/0.0035 = 2.002 = 2 (gets previous (1)mark also)Q(IV) or Q4+