How many grams of iron metal do you expect to be produced when 305 grams of a 75.5 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)2 (aq)

Question

How many grams of iron metal do you expect to be produced when 305 grams of a 75.5 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)2 (aq)

anonymous · Answer

(305 g) x (0.755) / (179.8548 g Fe(NO3)2/mol) x (3 mol Fe / 3 mol Fe(NO3)2) x 
(55.8452 g Fe/mol) = 71.5 g Fe