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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
Answer:
hola no ablo inalgun moderador puede acabar a este otro este me borró respuesta que no debió borrar
Answer:
I'm sorry, you didn't post the pictures.
