Please help?? I don't understand!
How many grams of iron metal do you expect to be produced when 245 grams of an 80.5 percent by mass iron (II) nitrate solution reacts with excess aluminum metal? Show all of the work needed to solve this problem.

2 Al (s) + 3 Fe(NO3)2 (aq) yields 3 Fe (s) + 2 Al(NO3)3 (aq)

Question

Please help?? I don't understand!
How many grams of iron metal do you expect to be produced when 245 grams of an 80.5 percent by mass iron (II) nitrate solution reacts with excess aluminum metal? Show all of the work needed to solve this problem. 

2 Al (s) + 3 Fe(NO3)2 (aq)  yields 3 Fe (s) + 2 Al(NO3)3 (aq)

anonymous · Answer

This is what i have so far
245 g Fe(NO3)2 x (80.5 g Fe(NO3)2 solute/ 100 g Fe(NO3)2) x ( 1 mol Fe(NO3)2/ 179.85 g Fe(NO3)2) x (1 mol Al/2 mol Fe(NO3)2) x (26.98 g Al/1 mol Al)