An iron-alloy nail (2.41g) is dissolved in 100cm3 acid. 10cm3 portions of this solution are titrated with KMnO4 (0.02M) and 9.80cm3 of KMnO4 was needed to react with iron solution. What % of iron by mass is in the nail?

This is a common example of a titration calculation which can be broken down into several key steps:1) Write a balanced equation for the redox reaction:MnO4 - + 8H+ + 5Fe2+ --> Mn2+ + 4H2O + 5Fe3+(This equation can either be learnt or derived)2) Work out moles of KMnO4 using equation: moles = conc x vol moles KMnO4 = 0.02 x (9.8/1000) = 0.000196 mol3) Using the moles of KMnO4, work out the moles of Fe2+ in 10 cm3 portionUsing the equation in Step 1, the ratio of KMnO4 to Fe2+ is 1:5, so we multiply our answer in Step 2 by 5.moles Fe2+ (in 10 cm3) = 0.000196 x 5 = 0.000980 mol4) Then work out the total Fe2+ moles in 100 cm3 solution Scale up our answer in Step 3 by 10 for the total solution Moles Fe2+ (in 100 cm3) = 0.00980 mol5) Work out the mass of Iron in the 100cm3 solution Using the equation mass = moles x Mr Mass Fe (in 100cm3 solution) = 0.00980 mol x 55.8 g/mol = 0.547 g6) To work out the mass %, divide the mass calculated in Step 5 but the total mass of the nail Mass % = 0.547/2.41 - 22.6%