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.

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)  3 Fe (s) + 2 Al(NO3)2 (aq) (4 points)

DrBob222 · Answer

You need to learn how to write arrows. --> or ==> or >>> will do it. Technically, the equation you wrote was incorrect since it doesn't separate the reactants from the products.
305 g sample Fe(NO3)2 has how much Fe(NO3)2? That's 305 x 0.755 = ?

mols Fe(NO3)2 = grams/molar mass
Use the coefficients in the balanced equation to convert mols Fe(NO3)2 to mols Fe.
Now convert mols Fe to grams Fe. g = mols x atomic mass Fe.