Ba(NO3)2 + Na2SO4 ==> BaSO4 + 2NaNO3
mols Ba(NO3)2 = M x L = ?
Using the coefficients in the balanced equation, convert mols Ba(NO3)2 to mols Na2SO4.
Then M Na2SO4 = mols Na2SO4/L Na2SO4. You know mols and M, solve for L and convert to mL.
What volume (in mL) of 0.55 M Na2SO4 solution is needed to precipitate all the barium, as BaSO4(s), from 15.5 mL of 0.25 M Ba(NO3)2 solution?
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