How many grams of iron metal do you expect to be produced when 265 grams of an 84.5 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) 3 Fe (s) + 2 Al(NO3)2 (aq)

Question

How many grams of iron metal do you expect to be produced when 265 grams of an 84.5 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)   3 Fe (s) + 2 Al(NO3)2 (aq)

anonymous · Answer

Remember grams - moles - moles - grams. 265 x 84.5 = 224g.   And then 224g Fe(NO3)2 x (1 mol Fe(NO3)2 / 179.8 g Fe(NO3)2) x (1 mol Fe / 1 mol Fe(NO3)2) x (55.8g / 1 mol Fe) = 69.5 g Fe