Answer:
319.2 g of iron metal
Explanation:
We'll begin by writing the balanced equation for the reaction between iron (Fe) and oxygen (O₂) to produce ferric oxide (Fe₂O₃). This is illustrated below:
4Fe + 3O₂ —> 2Fe₂O₃
Next, we shall determine the mass of Fe that reacted and the mass of Fe₂O₃ produced from the balanced equation. This can be obtained as follow:
Molar mass of Fe = 56 g/mol
Mass of Fe from the balanced equation = 4 × 56 = 224 g
Molar mass of Fe₂O₃ = (56×2) + (16×3)
= 112 + 48
= 160 g/mol
Mass of Fe₂O₃ from the balanced equation = 2 × 160 = 320 g
SUMMARY:
From the balanced equation above,
224 g of Fe reacted to produce 320 g of Fe₂O₃.
Finally, we shall determine the mass of iron metal, Fe, required to produce 456 g of ferric oxide, Fe₂O₃. This can be obtained as follow:
From the balanced equation above,
224 g of Fe reacted to produce 320 g of Fe₂O₃.
Therefore, Xg of Fe will react to produce 456 g of Fe₂O₃ i.e
Xg of Fe = (224 × 456)/320
Xg of Fe = 319.2 g
Thus, 319.2 g of iron metal, Fe is required to produce 456 g of ferric oxide, Fe₂O₃.
