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Moles Q: Barium sulfate can be prepared in the following precipitation reaction...

1451939738210-200064831.jpg

I started by using the n=cv equation to work out the moles for the equation mass=mr x moles and mr is also easy to get. However what to do next is what I am stuck on, help?
Reply 1
Avatar for Ben :)
Ben :)
2
Moles of BaCl2: (25/1000) x 0.1 = 0.0025 mol

You can see from the equation that there is a 1:1 ratio of moles between BaCl2 : BaSO4 so:

Moles of BaSO4: 0.0025 mol
Mr of BaSO4: 233.4

Mass = Moles x Mr
0.0025 x 233.4
0.5835g

As there is an 80% yield:

0.5835 x 0.8 = 0.4668g
0.47g so B!

😃
Original post by Ben :)
Moles of BaCl2: (25/1000) x 0.1 = 0.0025 mol

You can see from the equation that there is a 1:1 ratio of moles between BaCl2 : BaSO4 so:

Moles of BaSO4: 0.0025 mol
Mr of BaSO4: 233.4

Mass = Moles x Mr
0.0025 x 233.4
0.5835g

As there is an 80% yield:

0.5835 x 0.8 = 0.4668g
0.47g so B!

😃


I see, so the information given about Na2SO4 was not needed?
Thank you very much! 😊
Reply 3
Avatar for Ben :)
Ben :)
2
Original post by tessa.lin
I see, so the information given about Na2SO4 was not needed?
Thank you very much! 😊


No, that was just extra info that they put in to confuse you I think! I haven't checked but if you follow the same method with Na2SO4 then it should give you the same answer ☺️!

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