Because of this, railroads use the reaction to provide molten steel to weld steel rails together when laying track. When 23.00g of iron(III) oxide reacts with aluminum metal, how many moles of molten iron (Fe) are formed

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.