Answer: 88.2 g
Solution:
1) Chemical equation:
<span>2Al (s) + 3Fe(NO3)2 (aq) → 3Fe (s) + 2Al(NO3)3 (aq)
2) Theoretical molar ratios
2 mol Al : 3 mol Fe(NO3)2 : 3 mol Fe : 2 mol Al(NO3)3
3) Starting mass of pure iron nitrate
% = (mass of iron nitrate / mass of solution) * 100 = 87.5
=> mass of iron nitrate = 87.5 * mass of solution / 100
mass of solution = 325 g
=> mass of iron nitrate = 87.5 * 325 g / 100 = 284.375 g
4) moles of iron nitrate
moles = mass in grams / molar mass
molar mass of Fe(NO3)2 = 179.85 g/mol
moles = 284.375 g/ 179.85 g/mol = 1.58 moles Fe(NO3)2
5) proportion:
x 3 mol Fe
--------------------------- = ----------------------
1.58 mol Fe(NO3)2 3 mol Fe(NO3)2
Clear x:
x = 1.58 mol Fe
6) Convert 1.58 mol Fe into grams
mass = number of moles * atomic mass
atomic mass of iron = 55.845 g / mol
mass = 1.58 moles * 55.845 g/mol = 88.24 g
Rounded to 3 significant figures: 88.2 grams of Fe.
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