Answer
There is a total of $0.80$ mole $Al^{3+}$ in 0.40 mole $Al_2(SO_4)_3$
Work Step by Step
1. Identify the conversion factor for $Al_2(SO_4)_3$ moles and $Al^{3+}$ moles.
- Each $Al_2(SO_4)_3$ molecule has 2 aluminum ions ($Al^{3+}$). Therefore:
$\frac{1mole(Al_2(SO_4)_3)}{2moles(Al^{3+})}$ and $\frac{2moles(Al^{3+})}{1mole(Al_2(SO_4)_3)}$
2. Calculate the amount of $Al^{3+}$ (in moles), in $0.40$ mole of $Al_2(SO_4)_3$:
$0.40 mole(Al_2(SO_4)_3) \times \frac{2moles(Al^{3+})}{1mole(Al_2(SO_4)_3)} = 0.80 $ mole $ (Al^{3+})$
