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 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)

Procedure: determine the number of moles of iron (II) nitrate in the solution and use the ratios given by the chemical equation.

1) moles of iron (II) nitrate in the solution

grams of solute: 87.5% * 325 g = 284.375 g

Convert to moles by dividing by the molar mass of Fe [NO3]2

molar mass of Fe [NO3]2 = 179.9 g / mol ... (you can either calculate it from the atomics masses of the elements, or find it directly in Internet)

number of moles = 284.375 g / 179.9 g/mol = 1.58 mol

2) Ratios

From the balanced chemical equation 3 moles of Fe (NO3) 2 produces 3 moles of Fe, then 1.58 mol of Fe (NO3) 2 will produce 1.58 mol of Fe.

3) Convert 1.58 mol of Fe by multiplying by its atomic mass:

1.58 mol * 55.85 g/mol = 88.24 g of Fe

Answer: 88.24 g