Asked by Sadaf
Given the following balanced equation, determine the mass of iron metal that would be produced from the complete reaction of 94.7g of aluminum with excess iron II oxide.
2 Al(s) + 3FeO(aq) → 3Fe(s) + Al2O(aq)
2 Al(s) + 3FeO(aq) → 3Fe(s) + Al2O(aq)
Answers
Answered by
DrBob222
The equation you wrote is not correct. It should be Al2O3(s). Also, I don't know how you make FeO and Al2O3 in aqueous solutions.
2Al + 3FeO → 3Fe + Al2O3
mols Al = g/atomic mass = 94.7g/27 = 3.51. Convert to mols Fe this way.
3.51 mols Al x (3 mols Fe/2 mol Al) = 5.26 mols Fe produced.
Convert to g Fe this way. grams Fe = mols Fe x atomic mass Fe = ?
2Al + 3FeO → 3Fe + Al2O3
mols Al = g/atomic mass = 94.7g/27 = 3.51. Convert to mols Fe this way.
3.51 mols Al x (3 mols Fe/2 mol Al) = 5.26 mols Fe produced.
Convert to g Fe this way. grams Fe = mols Fe x atomic mass Fe = ?
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