How do you find the MAXIMUM mass of precipitate? :/
How do you find maximum mass of precipitate? 
What i did so far was that I worked out the moles:
100/1000 x 0.500= 0.05 moles
Now when it comes to Mr, is it the Mr of silver nitrate or calcium chloride?
I have tried both ways but my answer didn't match any of the options given
?
What i did so far was that I worked out the moles:
100/1000 x 0.500= 0.05 moles
•
moles x Mr = Mass
Now when it comes to Mr, is it the Mr of silver nitrate or calcium chloride?
I have tried both ways but my answer didn't match any of the options given
?
is the answer 8.95g?
Original post by huggleberry247
is the answer 8.95g?
nope! That's what I thought but it's 7.17 g ?!
@Adorable98
Moles=Mass/Mr
AgCl is the ppt, think of the silver nitrate test for chloride ions, white ppt is produced.
So 100/1000 x 0.05
ANS^ x (107.9+35.5) = 7.17g
Moles=Mass/Mr
AgCl is the ppt, think of the silver nitrate test for chloride ions, white ppt is produced.
So 100/1000 x 0.05
ANS^ x (107.9+35.5) = 7.17g
(edited 7 years ago)
Original post by Engineerrookie
Remember moles = mass / Mr
AgCl is the ppt, think of the silver nitrate test for chloride ions, white ppt is produced.
So 100/1000 x 0.05
ANS^ x (107.9+35.5) = 7.17g
AgCl is the ppt, think of the silver nitrate test for chloride ions, white ppt is produced.
So 100/1000 x 0.05
ANS^ x (107.9+35.5) = 7.17g
Oh I see!! Thank you!!
Original post by Adorable98
Oh I see!! Thank you!!
No problem! Good luck tomorrow!
Original post by Engineerrookie
No problem! Good luck tomorrow!
Thank you!!
Original post by Sakhib
From where did you find 107.9 and including 35.5 ??
107.9 is the molar mass of Ag, 35.5 is the molar mass of Cl. Add them together to work our the Mr of AgCl, which is needed to convert into mass in the final step.
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