Question

A chemist burns 160.0 g of Al in excess air to produce aluminum oxide, Al2O3. She produces 260.0 g of solid aluminum oxide. Write a balanced equation for the reaction. Determine the theoretical yield of Al2O3. Determine the percent yield.

Answer 1

Answer:

The balanced reaction that takes place when aluminium burns in excess of air:

4 Al   +  3 O₂ (air)   →  2 Al₂O₃

Mass of Aluminium burnt = 160 g

Mass of Al₂O₃ formed = 260 g (solid)

Therefor to account for the product, we can use the balancing along with the molar masses of the compounds formed,

Hence,  

4 moles of Al needs 3 moles of O₂ ,

Molar masses on reactant side  = 27 x 4 + 3 x 32 = 108 + 96 = 204 g

Molar masses on product side = 2 x ( 27 x 2 + 32 x 3 ) = 300 g

Applying, unitary method,

For 204 g of Al and air we get 300 g of Al₂O₃

Therefore, for (160 + x)  g of Al and air we get 260 g of Al₂O₃

204 / 160 + x  = 300 / 260

x = 16.8 g of O₂ is used (active part of air)

Answer 2

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