Rust has the formula Fe₂O₃,and it's made from iron metal reacting with oxygen over time.Assume that you had a 2.0-g block of pure iron metal that fully oxidized to rust and the mass of that product was 6.5 g.What must have been the mass of oxygen that initially reacted with pure iron?

Question

Quizplus · Accepted Answer

The formula for rust shows that for every mole of iron that reacts, 13 moles of oxygen are required. We can use this ratio to calculate the mass of oxygen that reacted with the 2.0 g block of iron.

First, we need to convert the masses to moles using the molar masses:
- Iron: 2.0 g / 55.85 g/mol = 0.0357 mol
- Rust: 6.5 g / (55.85 g/mol + 32.06 g/mol + 12 x 10.81 g/mol) = 0.0220 mol

Next, we can use the mole ratio from the formula to find the moles of oxygen:
- 0.0357 mol iron x 13 mol oxygen / 1 mol iron = 0.4641 mol oxygen

Finally, we can convert the moles of oxygen back to a mass:
- Oxygen: 0.4641 mol x 32.06 g/mol = 14.88 g

Therefore, the initial mass of oxygen that reacted with the 2.0 g block of iron was:
- Mass of oxygen = 14.88 g - Mass of oxygen in rust = 14.88 g - 10.00 g = 4.88 g
So the best choice is C.

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Rust Has the Formula Fe₂O₃,and It's Made from Iron Metal