Question

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?

Answer 1

So let's start with the chemical equation (let's let X represent the unknown element)
BaX+Na2SO4 --> BaSO4+NaX

Answer 2

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

Answer 3

n=11.21/233.4=?
@dahlinal

Answer 4

@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

Answer 5

n= 0.0480

Answer 6

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

Answer 7

Molar mass= 10/0.048

Answer 8

208.33

Answer 9

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

Answer 10

Molar mass of barium is 137.33g/mol

Answer 11

71

Answer 12

Perfect so that is the molar mass but we have to look back at the chemical formula
BaX+Na2SO4 --> BaSO4+Na2X

Answer 13

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

Answer 14

35.5

Answer 15

Yep and there is an element that has that molar mass

Answer 16

Cl

Answer 17

Also the actual chemical formula is
BaX2+Na2SO4 --> BaSO4+2NaX

Answer 18

And yes it is chlorine as your unknown

Answer 19

thank you!

Answer 20

No problem! Happy to help!