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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.

Respuesta :

scme1702
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


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