Answer
There is a total of 4.5 moles of $SO{_4}^{2-}$ in 1.5 mole of $Al_2(SO_4)_3$.
Work Step by Step
1. Identify the conversion factor for $Al_2(SO_4)_3$ moles and $SO{_4}^{2-}$ moles.
- Each $Al_2(SO_4)_3$ molecule has 3 $SO{_4}^{2-}$. Therefore:
$\frac{1mole(Al_2(SO_4)_3)}{3moles(SO{_4}^{2-})}$ and $\frac{3moles(SO{_4}^{2-})}{1mole(Al_2(SO_4)_3)}$
2. Calculate the amount of $SO{_4}^{2-}$ (in moles), in $1.5$ moles of $Al_2(SO_4)_3$:
$1.5 moles(Al_2(SO_4)_3) \times \frac{3moles(SO{_4}^{2-})}{1mole(Al_2(SO_4)_3)} = 4.5 $ moles $ (SO{_4}^{2-})$