Answer:
28.9%
Explanation:
Let's consider the following balanced equation.
2 FeS₂ + 11/2 O₂ ⇒ Fe₂O₃ + 4 SO₂
We can establish the following relations:
- The molar mass of Fe₂O₃ is 159.6 g/mol
- 1 mole of Fe₂O₃ is produced per 2 moles of FeS₂
- 1 mole of Fe is in 1 mole of FeS₂
- The molar mass of Fe is 55.84 g/mol
The amount of Fe in the sample that produced 0.516 g of Fe₂O₃ is:

The percent of Fe in 1.25 g of the ore is:

When the concentration of a reactant is increased, the chemical equilibrium will shift towards the products. More product is formed and the concentration of the reactants decreases as the concentration of the products increases.