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 react with excess aluminum metal? show all work needed to solve this problem.

The mass of Fe(NO3)2 you have is 245 g * 0.805 = 197.23 g 

The molar mass of Fe(NO3)2 is 179.85 g/mole so you have 1.0966 moles 

From the balanced equation, that is the number of moles of Fe you make since both Fe(NO3)2 and Fe have a 3 in front of them. 

The mass of Fe is 1.0966 moles * 55.85 g/mole (atomic mass of iron) = 61.246 g. From the information you are given, your answer should be in 3 sig figs so 61.2 g 

Moles of HCl = 8.2 L * 3.5 mole/L = 28.7 moles 

From the balanced equation, you need 1/2 the moles of zinc as you have of HCl so you need 14.35 moles. 

To get the mass of zinc, multiply moles of zinc by its atomic mass = 938.2 g. From the information given, your answer should be in 2 sig figs so I would report it as 9.4 * 10^2 g, that clearly shows 2 sig figs.