Please help me on how to work this problem, im not good with this sort of thing?

What mass of barium sulfate can be produced when 100.0 mL of a 0.100 M solution of barium chloride is mixed with 100.0 mL of a 0.100 M solution of iron(III) sulfate?Please help me on how to work this problem, im not good with this sort of thing?
First balance the reaction equation:


3 BaCl2 + Fe2(SO4)3 --%26gt; 3 BaSO4 + 6 FeCl3





Since it take 3 times as much (per mole) of BaCl2 to react with the Iron (III) Sulfate, only one third of the Iron solution will be used to react with all of the Barium chloride solution.





100 mL = 0.1 L


0.100 M means 0.1 moles per Liter


0.1 moles per Liter * 0.1 L = 0.01 moles of Barium chloride reacted to give 0.01 moles of Barium sulfate.





Barium sulfate has 233.4 grams per mole (add the atomic masses of Ba + S + 4*O)





0.01 moles * 233.4 g/mol = 2.334 gramsPlease help me on how to work this problem, im not good with this sort of thing?
100ml x .1M = 10mmol barium sulfate.





10mmol x molucular weight of barium sulfate (add together the weight of Ba, S, O,O,O,O) = ur answer in milligrams