Question

calculate the number of aluminum ions in 0.056g of aluminum oxide (Al2 O3)?

Answer 1

Al O3 --> Al+3 + O-2
molecular mass of Al2O3 is (27×2 + 16×3) =102 units.
that means 102 grams of Al2O3 produces 54 grams of Al.
here we have 0.056g of Al2O3. So it produces 0.028g of Al.

1 mole of Al is producing here. so it has 6.044×10^23 molecules.

Answer 2

☺ Hello mate__ ❤

◾◾here is your answer...

1 mole of Al2O3 = 2 × 27 + 3 × 16 = 102g

i.e.,

102g of Al2O3 = 6.022 × 10^23 molecules of Al2O3

Then, 0.051 g of Al2O3 contains

=(6.022×10^23/102) ×0.056 molecules of Al2O3

= 3.301 × 10^20 molecules of Al2O3

The no. Al^3+ in one molecules of Al2O3 is 2.

Here, Al^3+ = aluminium ion .

Therefore, The number of Al^3+ present in 3.11 × 10^20 molecules (0.056g) of Al2O3

= 2 × 3.301 × 10^20

= 6.612 × 10^20

I hope, this will help you.

Thank you______❤

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