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posted by  magnolia on 1/30/2008 10:34:18 PM  |  status: Live  

chemistry

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A sample of iron chloride weighing 0.300 g was dissolved in water and the solution was treated with AgNO3 to precipitate the chloride as AgCl. After precipitation was complete, the AgCl was filtered, dried, and found to weigh 0.678 g. Determine the empirical formula of the iron chloride.
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posted by Sparky the LabRat on 1/30/2008 11:24:03 PM  |  status: Live
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"thank u"
Response Details:
We know an amount of iron chloride wasw added to water and dissolved.  AgCl was added in excess to react with all the Cl atoms from the iron chloride solution to form AgCl.
The reaction we're interested in is:

FeClx + AgNO3 --> (Fe+3) + (Ag +1) + AgCl

There was .678 g of AgCl precipitated, using the molecular weight of AgCl we get the moles of AgCl = 4.731*10^-3 moles of AgCl.

Using the emperical formula we can see there is 1 mole of Ag and 1 mole of Cl in AgCl.

Thus there are 4.73*10-3 moles Cl in the AgCl precipitate.  Which is equal to .168 g of Cl that was precipitated out of the iron chloride solution.

Convert  moles Cl to g of Cl.  = .1677 g Cl.  Thus  .1677 g of Cl  were in the initial  iron chloride solution. 

Subtract g of Cl from the initial mass of iron chloride to get mass of iron = .3 g - .1677 g = .132 g Fe

Convert mass of iron to moles of iron = 2.37*10^-3 moles

Find the ration of moles of Fe to moles of Cl.  2.37*10^-3 moles Fe = x * (4.73*10^-3 moles Cl)

Moles of Fe to moles of Cl        Fe:Cl = 1:2

==> FeCl2 is the emperical formula of the starting iron compound

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posted by neha2008 on 1/31/2008 1:04:25 AM  |  status: Live
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magnolia's comment:
" thank u"
Response Details:
Mercury and bromine will react with each other to produce mercury(II) bromide.

Hg(l) + Br2(l) HgBr2(s)

(a) What mass of HgBr2 is produced from the reaction of 9.67 g Hg and 11.9 g Br2?
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What mass of which reactant is left unreacted?
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(b) What mass of HgBr2 is produced from the reaction of 6.09 mL of mercury (density = 13.6 g/mL) and 6.09 mL bromine (density = 3.10 g/mL)?
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can you help me find the first two parts plz which is half of with X by em...!!!
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