Calculate the mass of the excess reactant remaining in the crucible

Question

a mixture of 20.0 g of iron(ii) oxide ad 10.0 g of aluminum metal is placed in a crucible and heated in a high temp oven, where a reduction of the oxide occurs. Calculate the mass of the excess reactant remaining in the crucible

Answer 1

Iron(II) oxide = FeO

Aluminum = Al

3FeO + 2Al ==> Al₂O₃ + 3Fe ... balanced equation

Find limiting reactant

One way is to simply divide the moles of each reactant by the corresponding coefficient in the balanced equation and whichever result is the smallest, that represents the limiting reactant

For FeO we have 20.0 g FeO x 1 mol FeO / 71.8 g = 0.279 mol FeO (÷3->0.093)

For Al we have 10.0 g Al x 1 mol Al / 26.98 g = 0.371 mol Al (÷2->0.185)

Since 0.093 is less than 0.185, FeO is limiting

Thus Al is in excess.

How much Al remains? First, find out how many mols of Al were used up:

0.279 mol FeO x 2 mol Al / 3 mol FeO = 0.186 mols Al used up

Next, find mols of Al remaining:

0.371 mol - 0.186 mol = 0.185 mol Al remaining

Finally, convert this to mass (g):

0.185 mol Al x 26.98 g / mol = 4.99 g Al remaining