The iron solution was titrated with 41.08 mL of a 0.458 M solution of Ce+ according to the balanced chemical equation below:Ce4+ (+) Fe2+ ---------gt; Ce3+ (+) Fe3+Calculate the mass of iron in the original ore sample
The Ce and Fe react in a 1:1 ratio so we need to find the number of moles of Ce ions and this will equal the number of moles of Iron. Moles of Ce4+ Vol of solution (litres) x Molarity (moles/l) 0.04108 litres x 0.458 M 0.01881 Moles Mass of Fe No of moles x molar (in this case atomic) mass 0.01881 x 55.85g 1.051 g
The Ce and Fe react in a 1:1 ratio so we need to find the number of moles of Ce ions and this will equal the number of moles of Iron. Moles of Ce4+ Vol of solution (litres) x Molarity (moles/l) 0.04108 litres x 0.458 M 0.01881 Moles Mass of Fe No of moles x molar (in this case atomic) mass 0.01881 x 55.85g 1.051 g
