91Ó°ÊÓ

We need to find moles of $\mathrm{O}$in$3$moles of Aluminum sulfate .

Using equalities and conversion factor using subscripts:

One mole of ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$= $12$moles of $\mathrm{O}$

Conversion Factor=$\frac{12mole\mathrm{O}}{1mole{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3}$

Moles of role="math" localid="1652693859212" $\mathrm{O}$in role="math" localid="1652693863716" $3.00$moles of aluminum sulfate=$$\begin{gathered} \frac{12mole\mathrm{O}}{1mole{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3}×3mole{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3 \\ =36mole\mathrm{O} \end{gathered}$$

Hence, $36$moles of $\mathrm{O}$present in $3.00$moles aluminum sulfate.

We need to find moles of ${\mathrm{Al}}^{3+}$in $0.40$moles of Aluminum sulfate .

Using equalities and conversion factor using subscripts:

One mole of ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$= $2$ moles of ${\mathrm{Al}}^{3+}$

Conversion Factor= $\frac{2mole{\mathrm{Al}}^{3+}}{1mole{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3}$

Moles of ${\mathrm{Al}}^{3+}$ions in $0.40$moles of aluminum sulfate= $$\begin{gathered} \frac{2mole{\mathrm{Al}}^{3+}}{1mole{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3}×0.40mole{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3 \\ =0.80mole{\mathrm{Al}}^{3+} \end{gathered}$$

Hence, $0.80$ moles of ${\mathrm{Al}}^{3+}$present in $0.40$moles aluminum sulfate.

We need to find moles of ${{\mathrm{SO}}_4}^{2-}$ionsin $1.5$moles of Aluminum sulfate .

Using equalities and conversion factor using subscripts:

One mole of ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$= $3$ moles of role="math" localid="1652694756447" ${{\mathrm{SO}}_4}^{2-}$

Conversion Factor=role="math" localid="1652694763246" $\frac{3mole{{\mathrm{SO}}_4}^{2-}}{1mole{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3}$

Moles of ions in moles of aluminum sulfate=role="math" localid="1652694748574" $$\begin{gathered} \frac{3mole{{\mathrm{SO}}_4}^{2-}}{1mole{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3}×1.5mole{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3 \\ =4.5mole{{\mathrm{SO}}_4}^{2-} \end{gathered}$$

Hence, $4.5$ moles of role="math" localid="1652694735665" ${{\mathrm{SO}}_4}^{2-}$present in $1.5$moles aluminum sulfate.