a 10.00 gram sample of a soluble barium salt is treated with an excess of sodium sulfate to precipitate 11.21 grams BaSO4, (M= 233.4). which barium salt is it?

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a 10.00 gram sample of a soluble barium salt is treated with an excess of sodium sulfate to precipitate 11.21 grams BaSO4, (M= 233.4). which barium salt is it?

Chemistry
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So let's start with the chemical equation (let's let X represent the unknown element) BaX+Na2SO4 --> BaSO4+NaX
Next step is to determine the number of moles to do this we have the mass of our products of 11.21g we also know the molar mass of 233.4 n=mass/molar mass
n=11.21/233.4=? @dahlinal

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@dahlinal im more than happy to help but I can't just give you the answer I'm happy to walk you to it, but you're going to have to do some work here too
n= 0.0480
Yes so now we have to reverse it we have the moles and we have the weight and we will use it to determine the molar mass of the BaX
Molar mass= 10/0.048
208.33
Perfect so we know that the molar mass of the BaX=208.33 So if we subtract the molar mass of the barium we can find the molar mass of our unknown 208.33-Ba=x
Molar mass of barium is 137.33g/mol
71
Perfect so that is the molar mass but we have to look back at the chemical formula BaX+Na2SO4 --> BaSO4+Na2X
Ba has an oxidation number of 2+ so our unknown is actually X2 so we have to divide our mollar mass we found and divide that by 2
35.5
Yep and there is an element that has that molar mass
Cl
Also the actual chemical formula is BaX2+Na2SO4 --> BaSO4+2NaX
And yes it is chlorine as your unknown
thank you!
No problem! Happy to help!

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