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^+

Jul 18, 2017

(a) $3.70$ $\text{mol Al}$

(b) $5.55$ $\text{mol S}$

(c) $3.70$ ${\text{mol Al}}^{3 +}$

Explanation:

We're asked to find the number of moles of some species in $1.85$ ${\text{mol Al"_2"(SO"_4")}}_{3}$.

(a)

The moles of $\text{Al}$ atoms:

There are $2$ atoms of $\text{Al}$ per unit of ${\text{Al"_2"(SO"_4")}}_{3}$, so there are two moles of $\text{Al}$ per mole of ${\text{Al"_2"(SO"_4")}}_{3}$:

1.85cancel("mol Al"_2"(SO"_4")"_3)((2color(white)(l)"mol Al")/(1cancel("mol Al"_2"(SO"_4")"_3))) = 3.70 $\text{mol Al}$

(b)

The moles of $\text{S}$ atoms:

There are $3$ atoms of $\text{S}$ per units of ${\text{Al"_2"(SO"_4")}}_{3}$, so there are $3$ moles of $\text{S}$ per mole** of ${\text{Al"_2"(SO"_4")}}_{3}$:

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

(c)

The moles of ${\text{Al}}^{3 +}$ ions:

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