4 Fe + 3 O2 --> 2 Fe2O3
How many grams of Fe will react with 40.0 grams of O2?

93.33 g of Fe

Explanation:

The balanced equation for the reaction is given below:

4Fe + 3O₂ —> 2Fe₂O₃

Next, we shall determine the masses of Fe and O₂ that reacted 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 O₂ = 2 × 16 = 32 g/mol

Mass of O₂ from the balanced equation = 3 × 32 = 96 g

SUMMARY:

From the balanced equation above,

224 g of Fe reacted with 96 g of O₂.

Finally, we shall determine the mass of Fe required to react with 40 g of O₂. This can be obtained as follow:

From the balanced equation above,

224 g of Fe reacted with 96 g of O₂.

Therefore, Xg of Fe will react with 40 g of O₂ i.e

Xg of Fe = (224 × 40)/96

Xg of Fe = 93.33 g

Therefore, 93.33 g of Fe is required to react with 40 g of O₂.

3 hours because it is a decent distance.