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

The reaction will produce 71.5 g Fe .

Step 1. Write the balanced chemical equation

3Fe(NO_3)_2  + 2Al → 3Fe + 2Al(NO3)_3

Step 2. Calculate the mass of Fe(NO3)_2

Mass = 305 g solution × [75.5g Fe(NO_3)_2/100 g solution]

= 230.3 g Fe(NO_3)_2

Step 3. Calculate the moles of Fe(NO_3)_2

Moles of Fe(NO_3)_2

= 230.3 g Fe(NO_3)_2 × [1 mol Fe(NO_3)_2/179.85 g Fe(NO_3)_2]

= 1.280 mol Fe(NO_3)_2

Step 4. Calculate the moles of Fe

Moles of Fe = 1.280 mol Fe(NO_3)_2 × (3 mol Fe/3 mol Fe(NO_3)_2)

= 1.280 mol Fe

Step 5. Calculate the mass of Fe

Mass of Fe = 1.280 mol Fe × (55.845 g Fe/1 mol Fe) = 71.5 g Fe

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idk what the question is about