Question

a chemist burns 16.0 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 perfect yield

Answer 1

Answer:

86.035%, 52.94%

Explanation:

Please Note: The following solution is written by assuming 26 g of Al₂O₃ produced instead of 260 g. Make sure to check the data once in question and verify it.

DATA REQUIRED:

Molar mass of Aluminium = 27 g

Molar mass of Al₂O₃ = 2 x 27 + 3 x 16 = 102 g

The chemical reaction equation for the burning of Aluminium in presence of oxygen is:

4Al +3 O₂ ⇒ 2Al₂O₃

Given mass of Aluminium = 16 g

No of moles of Aluminium = 16/27 = 0.59 moles

From the chemical equation,

4 moles of Al produces 2 moles of Al₂O₃

⇒ ( 4 x 27 = 108 ) g of Al produces (2 x 102 = 204) g of Al₂O₃

⇒ 16 g of Al will produce = 30.22 gram

Now, according to question, mass of Al₂O₃ produced = 26.0 g

Theoretical yield =

(Mass of product produced actually/Mass of product produced theoretically) x 100

= (26/30.22) x 100

= 86.035%

To calculate Percent Yield, we need to know how much of Al fed is converted into Al₂O₃.

Again, reading the chemical reaction equation in reverse manner:

204 g of Al₂O₃ is produced by 108 g of Al

⇒ 16 g of Al₂O₃ will be produced by (108/204) x 16 = 8.47 g of Al

Hence, mass of Al put into reactor = 16 g

Mass of Al consumed (mass which is converted into product) = 8.47 g

Percent Yield =

(Mass of reactant converted into product/Mass of reactant fed into reactor) x 100

= (8.47/16) x 100

= 52.94 %