Question

calculate the number of Aluminium ions present in 2 g of Aluminium oxide
pls ans fast urgent
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Answer 1

Answer:

Mole of aluminium oxide (A1203) is

⇒ 2 x 27 + 3 x 16

Mole of aluminium oxide = 102 g

i.e., 102 g of Al2O3= 6.022 x 1023 molecules of A1203

Then, 0.051 g of A1203 contains = 6.022 x 1023 / (102 x 0.051 molecules)

= 3.011 x 1020 molecules of A1203

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 × 1020 molecules (0.051g) of aluminium oxide (Al203)

= 2 × 3.011 × 1020 = 6.022 × 1020

Explanation:

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Answer 2

Answer:

Molar mass of Aluminium Oxide (Al 2 O 3 ) = M = 102g.

$2 / 102 = N / 6.022 \times {10}^{23}$

$N = 118.07 \times {10}^{20}$

molecules.

1 molecule of Al 2 O 3 has 2 Al 3+ ions.

So, number of Al 3+ ions in 0.051 molecules of Al 2 O 3 is;

$118.07 \times 2 \times {10}^{20} = 236.14 \times {10}^{20}$