Question

Calculate the number of aluminium ions (Al3+) in 204g of alumina (Al2O3) please answer fast​

Answer 1

Solution

Mole of aluminium oxide (Al2O3) is

⇒ 2 x 27 + 3 x 16

⇒Mole of aluminium oxide = 102 g

i.e., 102 g of Al2O3= 6.022 x 10molecules of Al2O3

⇒Then, 0.051 g of Al2O3 contains = 6.0 x 10²³ / (102 x 0.051 molecules)

= 3.011 x 10²0 molecules of Al²O³

⇒The number of aluminium ions (Al3+) present in ⇒one molecule of aluminium oxide is 2.

⇒Therefore, the number of aluminium ions (Al3+) present in 3.11 × 10^20 molecules (0.051g) of aluminium oxide (Al 2 O 3)

= 2 × 3.011 × 10^20

= 6.022 × 10^20

Answer 2

Answer:

Mole of aluminium oxide (Al2O3) is

⇒ 2 x 27 + 3 x 16

⇒Mole of aluminium oxide = 102 g

i.e., 102 g of Al2O3= 6.022 x 10molecules of Al2O3

⇒Then, 0.051 g of Al2O3 contains = 6.0 x 10²³ / (102 x 0.051 molecules)

= 3.011 x 10²0 molecules of Al²O³

⇒The number of aluminium ions (Al3+) present in ⇒one molecule of aluminium oxide is 2.

⇒Therefore, the number of aluminium ions (Al3+) present in 3.11 × 10^20 molecules (0.051g) of aluminium oxide (Al 2 O 3)

= 2 × 3.011 × 10^20

= 6.022 × 10^20

Explanation:

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