help me please?

Question

What mass of barium sulfate can be produced when 200.0 mL of a 0.100 M solution of barium chloride is mixed with 200.0 mL of a 0.100 M solution of iron(III) sulfate?

___________ g

Answer 1

Here are the concepts.

You need to write a balanced equation. It would be

3BaCl3 + Fe2(SO4)3 == 3 BaSO4 + 2FeCl3.

Normally, you need to calculate the stoich. on both reactants to see which is limiting. But since you need 3 moles of Barium chloride for every one mole of Iron III sulfate, its clear that the Barium will get used up first. Since the 2 Barium compunds are in a 1:1 ratio, the stioch is actually quite easy. You will produce the equivalent of 200.0 ml of 0.100 M solution of BaSO4, or 0.05 moles. Multiply by the molecular mass to get the grams.