Answer to Question #155533 in Chemistry for N/A

Question

How many oxygen atoms are present in 4.90 mol in Al 203?

Answer 1

Solution:

Al₂O₃ → 2Al + 3O

According to the above scheme: n(Al₂O₃) = n(Al)/2 = n(O)/3

n(O) = 3 × n(Al₂O₃) = 3 × 4.90 mol = 14.7 mol

One mole of any substance contains 6.022×10²³ atoms/molecules.

Hence,

Number of O atoms = 14.7 mol × (6.022×10²³ atoms / 1 mol) = 8.85×10²⁴ O atoms

Number of O atoms = 8.85×10²⁴ O atoms

Answer: 8.85×10²⁴ oxygen atoms are present in 4.90 mol in Al₂O₃.