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Topic: Ba2+ + SO42- = BaSO4(s) Calculation  (Read 21404 times)

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Offline jiunshan

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Ba2+ + SO42- = BaSO4(s) Calculation
« on: February 07, 2006, 08:15:56 PM »
5.34g of salt of formula M2SO4 (where M is a metal) were dissolved in water. The sulphate ion was precipitated by adding excess barium chloride solution when 4.66g of barium sulphate (BaSO4) were obtained.
a) How many moles of sulphate ion were precipitated as barium sulphate?
b) How many moles of M2SO4 were in solution?

a) Barium and sulphate exist in a one-to-one ratio in the compound , so the number of moles of barium sulphate equals the number of moles of sulphate. Beside, how to get the exact value of moles? ???
« Last Edit: February 25, 2006, 11:47:53 PM by Mitch »

Offline Mitch

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Re:Calculation
« Reply #1 on: February 07, 2006, 09:09:31 PM »
Read line one in my signature.
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1. Start by writing a balanced chemical equation.
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Offline jiunshan

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Re:Calculation
« Reply #2 on: February 08, 2006, 02:27:29 AM »
oops..sorry!
Coz I'm confuse with this question so I donno how to write a equation for this question.
Can somebody help me?

Offline AWK

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Re:Calculation
« Reply #3 on: February 08, 2006, 02:51:24 AM »
Ba2+  + SO42- = BaSO4(s)

edit: typo corrected
« Last Edit: February 08, 2006, 03:11:04 AM by Borek »
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