Question

Calculate the number of aluminium was present in 0.051g of aluminium oxide(al2o3).

Answer 1

MOLAR MASS OF Al2O3 IS 102g

so moles of Al2O3  = given mass/molar mass

0.051/102=0.0005 moles

and each mole of Al2O3 contains 2 moles of aluminium so 0.0005 will contain

0.0005 *2 =0.001 moles of Al

as each moles contain 6.022* 10^23 atoms so 0.001 moles will contain

6.022*10^20 atoms of aluminium

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) ×0.051 molecules of Al2O3

= 3.011 × 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 (0.051g) of Al2O3

= 2 × 3.011 × 10^20

= 6.022 × 10^20

I hope, this will help you.

Thank you______❤

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