KTDMD gave a great solution, but just in case you are like me, who needs to see things really in pieces, this is kind of how I approached the problem. I broke it complete down into parts then put them together.

The question is asking for a percent of a whole, so the whole thing weighs 1.12g, and you want to find grams of Fe. So keep in mind you are looking for Fe2+ g first.

Part 1: So we have .05M of KMnO4 and 25mL of it. So .05M KMnO4 x .025L = # of moles of KMnO4.

Part 2: We have 5 moles of Fe2+ for every 1 mole of KMnO4, so in order to get the correct amount of Fe2+ we need to multiply the number of moles of KMnO4 by the number of moles of Fe2+. Therefore, we need to multiply by 5.

Part 3: We need GRAMS of Fe2+, so we should multiply the # of moles of Fe2+ by it's weight, 55.85.

Part 4: To make it into percent of whole you will take Grams of Fe2+ and divide by Grams of the whole thing = 1.12

Part 5: put it all together

(Moles of KMnO4) x (5 moles Fe/1 mole KMnO4) x (Fe2+ grams) / (Sample weight) x 100

(.05M x .025L) x (5/1) x (55.85g) / 1.12 x 100

Hope this helps!!