Question

Calculate the number of sulfate ions in 41.65 g of aluminum sulfate.

Answer 1

The number of sulfate ions : 2.2 x 10²³

Further explanation

Reaction

Al₂(SO₄)₃⇒2Al³⁺+3SO₄²⁻

mol Al₂(SO₄)₃(MW=342,15 g/mol) :

$\tt \dfrac{41.65}{342.15}=0.122$

mol of sulfate ions (SO₄²⁻)

$\tt 3\times 0.122=0.366$

1 mol = 6.02 x 10²³ particles (ions, molecules, atoms), so for 0.366 mol :

$\tt 0.366\times 6.02\times 10^{23}=2.20\times 10^{23}$