When a reaction mixture with a total volume of 4 L that contains 9.99 g of solid CaSO4 was stoichiometrically produced as per the balanced equation with 2.50 L of aqueous Ca2+, what molarity (M) of Ca2+ was required?
Al2(SO4)3(aq) + 3 CaBr2(aq) → 2 AlBr3(aq) + 3 CaSO4(s)
The Attempt at a Solution
Conversion of CaSO4 to mols = 1.36 mols
molar mass = 340.0 mol/L
After this, I don't know how to find the molarity of the Ca2+ ions...