A 1.85 mole sample of the ionic compound Al2(SO4)3 contains how many moles of the following? A) Al atoms B) S atoms C) Al3^+

1 Answer
Jul 18, 2017

(a) 3.70 "mol Al"

(b) 5.55 "mol S"

(c) 3.70 "mol Al"^(3+)

Explanation:

We're asked to find the number of moles of some species in 1.85 "mol Al"_2"(SO"_4")"_3.

(a)

The moles of "Al" atoms:

There are 2 atoms of "Al" per unit of "Al"_2"(SO"_4")"_3, so there are two moles of "Al" per mole of "Al"_2"(SO"_4")"_3:

1.85cancel("mol Al"_2"(SO"_4")"_3)((2color(white)(l)"mol Al")/(1cancel("mol Al"_2"(SO"_4")"_3))) = 3.70 "mol Al"

(b)

The moles of "S" atoms:

There are 3 atoms of "S" per units of "Al"_2"(SO"_4")"_3, so there are 3 moles of "S" per mole** of "Al"_2"(SO"_4")"_3:

1.85cancel("mol Al"_2"(SO"_4")"_3)((3color(white)(l)"mol S")/(1cancel("mol Al"_2"(SO"_4")"_3))) = 5.55 "mol S"

(c)

The moles of "Al"^(3+) ions:

The number of ions of "Al"^(3+) will be the same as the number of atoms of "Al", so the answer will be the same as that for (a) (assuming I understand the question correctly).