I really NEED help with a few chemistry problems thanks!!?

[Jina L]

A 50.0 mL aliquot of a sulfuric acid solution was treated with barium chloride and the resulting BaSO4 was isolated and weighed. If 0.667 g BaSO4 was obtained what is H2SO4's molarity?

H2SO4 + BaCl2 ---->BaSO4 + 2HCl

I know I start with the mL but i dont know where to go from there....
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which combination of 1 M solutions will produce a visible reaction when mixed?
a. H3PO4 and Na2SO3
b. Na2CO3 and KCl
c. H3PO4 and NaCl
d. H3PO4 and NaOH
*would i use the solubility rules to answer this?

[James]

You actually need to start with the weight of the BaSO4 created. Its molecular weight is 233.43 g/mol, so 0.667 g is 2.74 x 10^-3 mol. For that much to have been produced, the same number of moles of H2SO4 must have been present in the solution. 2.74 x 10^-3 mol / 50.0 mL = 2.74 x 10^-3 mol / 0.05 L = 0.055 mol/L or 0.055 M.

A visible reaction would likely be an acid-base reaction. H3PO4 is a weak acid, but 1M is relatively concentrated. NaOH is a very strong base and would react well with H3PO4.