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 reacts with excess aluminum metal? Show all of the work needed to solve this problem.

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

See your problem above with Mg and HCl. This is the same type of problem. The 80.5% is handled this way.

285g impure iron(II) nitrate x 0.805 = ? g pure iron(II) nitrate

245 x 0.8= 196/242=0.81

To calculate the number of grams of iron metal produced when 245 grams of an 80.5% by mass iron (II) nitrate solution reacts with excess aluminum metal, we need to follow a few steps:

Step 1: Determine the moles of iron (II) nitrate in the solution.
Given:
Mass of the iron (II) nitrate solution = 245 grams
Percentage by mass of iron (II) nitrate = 80.5%

To find the moles of iron (II) nitrate, we need to divide the mass of iron (II) nitrate by its molar mass. To do this, we should first find the molar mass of iron (II) nitrate:

Fe(NO3)2:
Iron (Fe) atomic mass = 55.85 g/mol
Nitrate (NO3) atomic mass = 62.01 g/mol
Molar mass of Fe(NO3)2 = (2 * 62.01) + 55.85 = 179.87 g/mol

Now, we can calculate the moles of iron (II) nitrate:
Moles of iron (II) nitrate = Mass of iron (II) nitrate / Molar mass of iron (II) nitrate
Moles of iron (II) nitrate = 245 g / 179.87 g/mol

Step 2: Use the stoichiometry of the balanced equation to relate the moles of iron (II) nitrate to moles of iron metal.
According to the balanced equation:
2 Al(s) + 3 Fe(NO3)2(aq) -> 3 Fe(s) + 2 Al(NO3)3(aq)
From the equation, we can see that the stoichiometric ratio between Fe(NO3)2 and Fe is 3:3.

Moles of iron = Moles of iron (II) nitrate * (3 moles of Fe / 3 moles of Fe(NO3)2)

Step 3: Convert moles of iron to grams.
We can convert moles of iron to grams using the molar mass of iron (Fe).

Molar mass of iron (Fe) = 55.85 g/mol

Grams of iron = Moles of iron * Molar mass of iron

Now, let's substitute the values and find the answer:
Moles of iron (II) nitrate = 245 g / 179.87 g/mol
Moles of iron = (245 g / 179.87 g/mol) * (3 moles of Fe / 3 moles of Fe(NO3)2)
Grams of iron = Moles of iron * Molar mass of iron

By following these steps and performing the calculations, we can determine the grams of iron metal produced in the reaction.