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Calculation question (1 Viewer)

babberz

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http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2012exams/pdf_doc/2012-hsc-exam-chemistry.pdf

how do you do the last multiple choice question.

I approached it in two methods, and got it right on one account (by selecting the nearest answer) and wrong on another account. What is the proper way to answer it.
Pb2+ + 2CL- ----> PbCl2(s)

the dried precipitate is PbCl2 so we calculate the moles of it
0.595/(207+70.9)
= approximately 0.002 moles

then we find mass of Pb
0.002x207
=0.443g
then to put it into g/L we divide by the volume
0.443/0.05
=8.86g/L
just use exact values all the way until the end to get 8.87
 
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When excess barium chloride solution was added to a solution by dissolving 2.51g of commercial lawn food in 250ml water, 1.03g barium sulfate was formed. The percentage sulfur in the lawn food is?
 

HeroicPandas

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When excess barium chloride solution was added to a solution by dissolving 2.51g of commercial lawn food in 250ml water, 1.03g barium sulfate was formed. The percentage sulfur in the lawn food is?
BaSO4 <--> Ba(2+) + SO4(2-)


n(BaSO4) = (1.03)/(137.3 + 32.07 + 64) (via n=m/M)


.: n(BaSO4) = 0.0044....mol

Now for every mole of barium sulfate, looking at its chemical formula, there is only 1 sulfur atom in it

Hence n(BaSO4) = n(S)

.: n(S) = 0.0044...mol

.: m(S) = 0.0044...x 32.07 = 0.1415....g


.: = 0.1415.../2.51 x 100 = 5.64% (w/w) (3sf)
 
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thank you. i don't have the solution but i did the same thing so i guess it would be right.
 

bedpotato

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There's another way to do it.

1. Find the percentage of sulfate - this comes to 16.89...%
2. For percentage of sulfur:
( molar mass of sulfur / molar mass of sulfate ) x percentage of sulfate
(32.07 / 96.07) x 16.89... = 5.64%
 

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