Fe2O3+2Al=Al2O3+2Fe

A welder has 1.873 _ 10^2 g Fe2O3 and 94.51 g Al in his welding kit. Which reactant will he run out of first? (options: Fe2O3 or Al ... it's Fe2O3).

How much of this reactant should he order to make sure he runs out of both reactants at the same time?

He should order ___g of the limiting reactant.

92.41g, this is the correct answer

We determine the limiting reactant by using the moles present in the equation and the actual moles.

According to equation, ratio of Fe₂O₃ : Al = 1 : 2

Actual moles of Fe₂O₃ = 187.3 / (56 x 2 + 16 x 3)

= 1.17

Actual moles of Al = 94.51 / 27

= 3.5

Fe₂O₃ is limiting. Fe₂O₃ required:

(moles Al) / 2 = 3.5/2 = 1.75

Moles to be added = 1.75 - 1.17

= 0.58

Mass to be added = moles x Mr

= 0.58 x (56 x 2 + 16 x 3)

= 92.8 grams