I need some simple Chemistry help on a problem.

[SOSTrooper]

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?

All I got is the equation setup

3 BaCl2 + Fe2(SO4)3 --> 3 BaSO4 + 2 FeCl3

Ba is 137.3g/mol
S is 32.07g/mol
O is 16.00g/mol

I'm lost from this point on; not sure how to set the math part up.

 

[vexingv]

you find out the number of moles you have from the molarity....
then use then in the ratios according to the equation
i didnt look at the numbers carefully, but double check to see if either could be a limiting reagent

 

[akodi]

find the limiting reagent then the rest should be easy peasy japanesy

 

[BigPoppa]

0.1 L x 0.1 M BaCl2 = 0.01 moles Ba and Cl2. Also, 0.01 moles Fe2 and S04. Mole ratio 3 moles BaCl2 to 1 mole Fe2(SO4)3. All the BaCl2 is consumed, 0.0033 moles Fe2(SO4)3 consumed (1/3 of the iron3 sulfate). Thus, 3 moles BaSO4 produced. Moles x molar mass = mass.