Answer
a. 9.0 moles S
b. 0.80 mole $Al^{3+}$
c. 4.5 moles $SO_4^{2-}$
Work Step by Step
a. Each $ Al_2(SO_4)_3 $ has 3 S atoms, thus:$$ 3.0 \space moles \space Al_2(SO_4)_3 \times \frac{ 3 \space moles \ S }{1 \space mole \space Al_2(SO_4)_3 } = 9.0 \space moles \space S $$
b. Each $ Al_2(SO_4)_3 $ has 2 $Al^{3+}$ ions, thus:$$ 0.40 \space mole \space Al_2(SO_4)_3 \times \frac{ 2 \space moles \ Al^{3+} }{1 \space mole \space Al_2(SO_4)_3 } = 0.80 \space mole \space Al^{3+} $$
c. - Each $ Al_2(SO_4)_3 $ has 3 $SO_4^{2-}$ ions, thus:$$ 1.5 \space moles \space Al_2(SO_4)_3 \times \frac{ 3 \space moles \ SO_4^{2-} }{1 \space mole \space Al_2(SO_4)_3 } = 4.5 \space moles \space SO_4^{2-} $$
