Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: importfan878 on November 22, 2007, 09:34:50 PM
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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?
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balanced equation?
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Yeah i got that far
H2S04 + Pb(C2H3O2)2 = PbSO4 + 2H(C2H3O2)
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okay good,
how about some stoichiometry then, if you can show us your working out, we can help you better
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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
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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?
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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.
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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
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Quick note, "G" does not mean "grams", "g" means "grams".
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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?
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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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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.
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you are correct, as Borek as confirmed for you :)
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alright so i have all part a done right. How do i do part b
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yo i am in zamitts class are you in honors chem???
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alright so i have all part a done right. How do i do part b
Use definition of Percent yield.
http://www.chem.purdue.edu/gchelp/gloss/percentyield.html
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B. What would be the percent yield of lead II sulfate if 15.2 G are recovered?
pls check that this is the value of PbSO4 recovered, compare with the theoretical yield that you just calculated, follow Padfoot's link for calculation help
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link didnt work but if its the same as theoretical yield i got 217.609 g. And yeah im in honers chem who r u
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link didnt work but if its the same as theoretical yield i got 217.609 g.
Percent yield is the actual yield/thoeretical yield * 100
Also:
pls check that this is the value of PbSO4 recovered
I havn't read the whole question, but it seems wrong
And yeah im in honers chem who r u
Padfoot, not wcfl ;)
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uhh what? and yeha that wrong its 9.31g
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No.
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do two seperate 3 step equations then take your calculated results and divide by 15.2 then multiply by 100. do this for both calculated results
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divide by 15.2 then multiply by 100.
15.2 is the ACTUAL yield, not the theoretical yield. So you've got in slightly around the wrong way here.
I doubt 15.2g is right in any case as the actual yield should'nt be greater that theoretical yield (from the posts above, the theoretical yield is 9.32g).