Answer:
0.288 mole of Fe.
Explanation:
We'll begin by calculating the number of mole present in 23.00g of iron(III) oxide (Fe2O3). This can be obtained as follow:
Mass of Fe2O3 = 23 g
Molar mass of Fe2O3 = (56×2) + (16×3)
= 112 + 48
= 160 g/mol
Mole of Fe2O3 =?
Mole = mass /Molar mass
Mole of Fe2O3 = 23 / 160
Mole of Fe2O3 = 0.144 mole
Next, we shall write the balanced equation for the reaction. This is given below:
2Al + Fe2O3 —> Al2O3 + 2Fe
From the balanced equation above,
1 mole of Fe2O3 reacted to produce 2 moles of Fe.
Finally, we shall determine the number of mole of Fe produced by the reaction of 23 g (i.e 0.144 mole) of Fe2O3. This can be obtained as illustrated below:
From the balanced equation above,
1 mole of Fe2O3 reacted to produce 2 moles of Fe.
Therefore, 0.144 mole of Fe2O3 will react to produce = 0.144 × 2 = 0.288 mole of Fe.
Thus, 0.288 mole of Fe is obtained from the reaction.
