Rust has the formula Fe2O3 and it’s made from iron metal reacting with oxygen over time.

Question

Rust has the formula fe2O3 and it’s made from iron metal reacting with oxygen over time. Assume you had a 2.0 g block of pure metal than fully oxidized to rust and the mass of that product was 6.5g. What must have been the mass of oxygen that initially reaction with pure iron?

Answer 1

First, we start with a balanced equation:

4Fe + 3O₂ => 2Fe₂O₃

From this reaction, we can gather the mole to mole ratios that we will need for the problem.

6.5 g Fe₂O₃ (1 mol Fe₂O₃/ 159.7 g Fe₂O₃) = 0.0407 mol Fe₂O₃ Now we will take this and multiply by the

mole to mole ration of Fe₂O₃ and O₂

0.0407 mol Fe₂O₃ (3 mol O₂/ 2 mol Fe₂O₃) = 0.0611 mol O₂ Now we will have to multiply by the

atomic mass of the Oxygen molecule, O₂.

0.0611 mol O₂ (32 g O₂/ 1 mol mol O₂) = 1.95 g O₂

To maintain significant figures, the mass of the oxygen would have been 2.0 grams.

Answer 2

1. Balanced equation 1st.. 2 Fe + 3 O₂ -----> 2 Fe₂O₃ Hmmmmmm Have you gotten to this (stoichiometry) yet? 2.0 g + x g = 6.5g If you doing this by conservation of mass: What is 6.5 g minus 2.0g..... Let me know if you need more info!