How do you calculate the percentage of sulfate content by volumetric analysis?

I've done the tests and all no worries, but im struggling to find the right procedure to calculate the amount of sulfate present in epsom salt. I added 15 g to 150 ml .1 mol HCL then titrated with a .1 mol BaCL soloution. I found the end point to be 22.6 ml. Any help would be awsome :)How do you calculate the percentage of sulfate content by volumetric analysis?
Break it down a bit

Epsom salt is MgSO4

So we are looking at

MgSO4 + HCl ==BaCl==%26gt; ?



So break the compounds down a bit



Mg^2+ + SO4^2- + H^1+ + Cl^1-



Flip them and you get

MgCl2 + H2SO4



So, now we know what you got at the end of the mixing the MgSO4 and HCl, both of which is soluble and aqueous.

The balanced equation is

MgSO4 + HCl ==BaCl==%26gt; MgCl2 + H2SO4



So now we need to figure out the theoretical yield of this for H2SO4

We have the weight in grams of MgSO4, we need the weight in grams of the .1 mol HCl

To find the weight of HCl in .1 molar HCl we can use the following

density(g/mL) multiplied by volume(ml) = the value for 100% HCl, multiply by .1 = weight of HCl in 1. mol HCl



So

(1.18 g/ml)(150 ml)(.1 mol) = 17.7 grams of HCl in 150mL of 0.1 molar solution.



Now, we've got our masses of each of our reactants, now we can find the limiting reagent.



15g MgSO4 (1 mol MgSO4/120.37g MgSO4) (1 mol H2SO4 / 1 mol MgSO4) (98.08g H2SO4 / 1 mol H2SO4) = 12.22 g H2SO4

17.7g HCl (1 mol HCl/36.46g HCl) (1 mol HCl/1 mol H2SO4) (98.08g H2SO4 / 1 mol H2SO4) = 47.61 g H2SO4



So MgSO4 is your limiting reagent and your ideal yield would 12.22g of H2SO4. To check your numbers we can convert the 12.22g of H2SO4 into ml by multiplying it by it's density (which is 1.84 g/ml)

So

12.22g x 1.84 g/ml = 22.48 ml H2SO4



If you factor in human error, you probably over measured something, but you should be pretty well set.

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