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.

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

92.41g, this is the correct answer