Question

If 1/2 moles of oxygen combine with Aluminimumto form Al2O3, then weight of Aluminium metal usedin the reaction is (Al=27) -​

Answer 1

Answer:

18g

Explanation:

Balanced reaction is

4Al+3O2=2Al2O3

3moles of oxygen oxidizes 4 moles of Al

1/2mole of oxygen oxidizes 4/6=2/3 moles of Al

1 mole of Al weighs 27g

2/3moles of Al weighs 2/3×27=18g

Answer 2

The weight of Aluminium metal used in the reaction is 18 g

Explanation:

The formation of Al₂O₃ is given by the equation below:

4Al + 3O₂ → 2Al₂O₃

Thus, from the above reaction, we can understand that 3 moles of oxygen combines with 4 moles of aluminium.

From question, 1/2 moles of oxygen combine with x moles of aluminium.

x = 1/2 × 4/3 = 4/6 = 2/3

From question, we know that the weight of the aluminium is 27 g.

Thus, the weight of aluminium in this reaction is:

⇒ W = 2/3 × 27

∴ W = 18 g