Question

How many grams of iron metal do you expect to be produced when 325 grams of an 87.5 percent by mass iron (II) nitrate solution react with excess aluminum metal? Show all of the work needed to solve this problem.
2Al (s) + 3Fe(NO3)2 (aq) yields 3Fe (s) + 2Al(NO3)3 (aq

Answer 1

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

Find the pure amount of iron nitrate solution:

325 g x 87.5% = 284.375 g

Convert 284.375g into mols by dividing by molar mass of the iron nitrate solution

Molar mass of iron nitrate =  55.85 g/mol + 2x14g/mol + 6x16g/mol = 179.85 g/mol

Moles of iron nitrate = 284.375 g / 179.85 g/mol = 1.58

Stoichiometry

3 mols of iron nitrate / 3 mols of iron = 1.58 mols of iron nitrate / x mols of iron

x = 1.58 mols of iron.

Convert that in grams by multiplying by the molar mass of iron

grams of iron = 1.58 mol x 55.85 g/mol = 88.24 g