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Topic: Ap CHem Hep #2  (Read 19066 times)

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Offline importfan878

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Ap CHem Hep #2
« on: November 22, 2007, 09:34:50 PM »
A. Solutions of sulfuria acid and lead II acetate react to form solid lead II sulfate and a solution of acetic acid. If 10.0 Grams of Sulfuria acid and 10 G of lead acetate are mixed, calculate number of grams of sulfuric acid, lead II acetate, Lead II sulfate and acetic acid present in mixture after complete.

B. What would be the percent yield of lead II sulfate if 15.2 G are recovered?

Offline LQ43

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Re: Ap CHem Hep #2
« Reply #1 on: November 22, 2007, 09:43:43 PM »
balanced equation?

Offline importfan878

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Re: Ap CHem Hep #2
« Reply #2 on: November 22, 2007, 09:49:37 PM »
Yeah i got that far

H2S04 + Pb(C2H3O2)2 = PbSO4 + 2H(C2H3O2)

Offline LQ43

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Re: Ap CHem Hep #2
« Reply #3 on: November 22, 2007, 10:01:02 PM »
okay good,

how about some stoichiometry then, if you can show us your working out, we can help you better

Offline importfan878

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Re: Ap CHem Hep #2
« Reply #4 on: November 22, 2007, 10:28:30 PM »
Not sure what stoichiometry is? Is it the "3 step prolblem". I was able to find that 9.13 G of Pbso4 will be created with the limiting compound being Pb(C2H302)2. 10 G could of H2S04 can make 30.933 G of PbSO4 provided enough Pb(C2H302)2.

I also found 10g of H2S04 can make 19.987 G of H(C2H3O2). It is limited by 10g of Pb(C2H3O2)2 producing 3.69 G of PbSO4

Offline LQ43

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Re: Ap CHem Hep #2
« Reply #5 on: November 22, 2007, 11:17:03 PM »


these calculations are known as stoichiometry, yes it is at least a "3 step problem"

recheck 
10g of Pb(C2H3O2)2 producing 3.69 G of PbSO4

I get 9.32g of PbSO4 from 10 g Pb(C2H3O2)2,

even so compare with the 30.933g PbSO4 from the 10 g of H2SO4, how much PbSO4 is really made?

which one Pb(C2H3O2)2 or H2SO4 gets used up?
use the one that gets used up to calculate for the true mass of HC2H3O2 produced

do you know how to find how much of the one that is left over?

Offline importfan878

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Re: Ap CHem Hep #2
« Reply #6 on: November 24, 2007, 11:30:56 AM »
alright i got that anwser too. 9.31 G of PbSO4 will be used.Pb(C2H3O2)2 is used up. That will created 3.69 G of H(C2H3O2). I don't know how to find how much is left over. Thats as far as i got.

Offline LQ43

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Re: Ap CHem Hep #2
« Reply #7 on: November 24, 2007, 01:01:10 PM »
good,

Write down what you know so far, it would be good to make a little table of what you started with, what reacts, what's left over

H2S04 + Pb(C2H3O2)2 = PbSO4 + 2H(C2H3O2)

Reactants:
10g Pb(C2H3O2)2 gets used up  - 0 grams after rxn
10 g H2SO4 - if its not used up, there is some left over.
a) to find Left over: do a similar calculation as before starting with 10g Pb(C2H3O2)2 to find how much H2SO4 will be used at the same time.
b) then  10g  H2SO4  starting amt  - H2SO4 used up = left over H2SO4

Products:
PbSO4 - 9.31 g already calculated
HC2H3O2: pls recheck your calculation, remember it is 2 HC2H2O2 : 1 Pb(C2H3O2)2 ratio

keep going, you'll get there

Offline enahs

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Re: Ap CHem Hep #2
« Reply #8 on: November 24, 2007, 01:50:11 PM »
Quick note, "G" does not mean "grams", "g" means "grams".


Offline importfan878

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Re: Ap CHem Hep #2
« Reply #9 on: November 25, 2007, 03:49:30 PM »
I checked my anwser and still got 3.69. I also got 19.187 owuld be able to made from h2s04 but i dont think that matters. And i found that 6.985g of h2s04 will be left. So i thnk i all still need to do is find how much h(c2h302) is made, am i right?

Offline Borek

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Re: Ap CHem Hep #2
« Reply #10 on: November 25, 2007, 04:37:55 PM »
3.692 g it is if you start with 10.00 g of lead acetate. Entered sulfuric acid amount is displayed in red - which means there was excess. 3.015 g of sulfuric acid was used, so there is 10-3.015=6.985 g left.

Seems to me A is almost done - you have not stated yet how much lead (II) acetate is left.
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Offline importfan878

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Re: Ap CHem Hep #2
« Reply #11 on: November 25, 2007, 05:11:04 PM »
Yeha i think i already did that earlier i foudn there was 6.985g of h2s04 excess. Also lead 2 acetate was used up so non left after reaction. I think all i need is h(C3h302) which i think is 3.69g, the part im confused on.

Offline LQ43

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Re: Ap CHem Hep #2
« Reply #12 on: November 25, 2007, 05:18:23 PM »
you are correct, as Borek as confirmed for you :)


Offline importfan878

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Re: Ap CHem Hep #2
« Reply #13 on: November 25, 2007, 06:20:23 PM »
alright so i have all part a done right. How do i do part b

Offline wcfl

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Re: Ap CHem Hep #2
« Reply #14 on: November 25, 2007, 06:25:25 PM »
yo i am in zamitts class are you in honors chem???

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