How many aluminum atoms are in 2.58 grams of aluminum

5.75x10^22 atoms

Explanation:

From a clear understanding of Avogadro's hypothesis, we discovered that 1mole of any substance contains 6.02x10^23 atoms. This implies that 1mole of Al also contains 6.02x10^23 atoms.

1mole of Al = 27g

If 27g of Al contains 6.02x10^23 atoms, then 2.58g of Al will contain = (2.58 x 6.02x10^23) / 27 = 5.75x10^22 atoms