Question

If 0.755 g of Al3+ dissolved in 960 mL of water reacts stoichiometrically according to the balanced equation, what volume (mL) of 0.0875 M aqueous Cl- is required?

Al2(SO4)3(aq) + 3 BaCl2(aq) → 2 AlCl3(aq) + 3 BaSO4(s)
It's late for me and easy to get confused. Check me out thoroughly.
0.755 g Al^+/molar mass Al x [1 mol Al2(SO4)3/2 mols Al^+3] = moles Al2(SO4)3.
That x 3/2 = moles BaCl2 which contains twice that much Cl^-.
Then moles Cl^-/L = M. Solve for L and multiply by 1000. Check my thinking. I'm off to bed. It's approaching 1:00 AM here.

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