Can anyone give me a hand with some homework?

[Indigo]

I was doing fairly well in chemistry this semester until some health problems kept me out of class for two weeks....now I am totally lost and desperately need help. I have 6 problems due tomorrow all like the following. I am usually pretty good at figuring this sort of thing out on my own, but I can not find a problem like this in my text that is broken down and done step by step. I have no idea where to even begin.

"Ordinary chalkboard is a solid mixture, with limestone (CaCO3) and gypsum (CaSO4) as its principal ingredients. Limestone dissolves in dilute HCL(aq) but gypsum dies not according to the following unbalanced reaction:

CaCO3(s) + HCL(aq) -----> CaCl2(aq) + CO2(g) + H2O(l)

(a.) If a 12.3g piece of chalk that is 69.7% CaCO3 is dissolved in excess HCL(aq), what mass of CO2(g) will be produced?

(b.) Determine the mass percent of CaCO3 in a 4.38g piece of chalk that yields 1.31.g CO2 when it reacts with excess HCL(aq)

I guess what I am looking for is someone to show me how this would be done step by step, then, hopefully, I can apply it to the other 5...

Any help here would be GREATLY appreciated...

[reject]

Are you female? Do I get e-sex if I do it?

[resmeplz]

Originally Posted by Indigo

I was doing fairly well in chemistry this semester until some health problems kept me out of class for two weeks....now I am totally lost and desperately need help. I have 6 problems due tomorrow all like the following. I am usually pretty good at figuring this sort of thing out on my own, but I can not find a problem like this in my text that is broken down and done step by step. I have no idea where to even begin.

"Ordinary chalkboard is a solid mixture, with limestone (CaCO3) and gypsum (CaSO4) as its principal ingredients. Limestone dissolves in dilute HCL(aq) but gypsum dies not according to the following unbalanced reaction:

CaCO3(s) + HCL(aq) -----> CaCl2(aq) + CO2(g) + H2O(l)

(a.) If a 12.3g piece of chalk that is 69.7% CaCO3 is dissolved in excess HCL(aq), what mass of CO2(g) will be produced?

(b.) Determine the mass percent of CaCO3 in a 4.38g piece of chalk that yields 1.31.g CO2 when it reacts with excess HCL(aq)

I guess what I am looking for is someone to show me how this would be done step by step, then, hopefully, I can apply it to the other 5...

Any help here would be GREATLY appreciated...

Balance the equation, figure out how much CaCO3 you have, then use mole ratios to figure out how much CaCO3 is converted to CO2.

For B you do it backwards, you know how much chalk and CO2 you have so figure out how much CaCO3 there is, then divide that by 4.38x100 to get the mass percent.

I tried to explain it the best I could ^^ if you still need help i can run and get some paper, just to lazy to walk outside to get it

[Indigo]

Well I got as far as balancing it and figuring out how much calcium carbonate....and that's about it. I am totally blown away how far I fell behind in such a short period (this is all rather embarrassing..). I'm still working on it, and I don't want to make you run outside, but if anyone else has anything to say I'll check back periodically.

Zarni, sorry but I am a guy....still, I am not above a lil e sex with a man if it gets this solved .

Thanks guys.

[resmeplz]

ok, figure out how many moles of CaCO3 you have by calculating the molecular weight of CaCO3 and set up a dim. analysis to convert the grams of CaCO3 to moles of CaCO3 then multiply that by the ratio of moles of CO2 to moles of CaCO3 (you use the co-efficients from your balanced equation) which will give you moles of CO2. from there you can convert to grams.

I'm not doing shit so if you still need help just ask.

EDIT: e-sex huh? wait, how old are you?

[Indigo]

lol res I was answering the post above yours...

I certainly appreciate the help you have offered so far and I came up with a number for part a...3.77g CO2 produced. Sound about right?

Onto part b....

Thanks again, there is no way I would have come up with an answer without your help.

[Michael Scott]

are you serious? This shit is easy. All you have to do is borrow and carry the "1"
retards.

[Indigo]

It's asshole comments like that that made me so self reliant in the first place.

Anyway, I got it now. Yet again, thank you res.

[resmeplz]

Yeah, no problem. Chemistry is kind of my thing so pm me anytime.

[atara]

We give you a hard time to encourage you to do as much as you can, not to scare you off.

>:3

[beaker]

sorry if I am a bit late with assisting you, I guess its never too late when it comes to knowledge so I"ll see if I can help you anyway.

The reaction between calcium carbonate and hydrochloric acid can be written two ways:
CaCO3 + 2 HCl → CaCl2 + H2CO3 and H2CO3 → H2O + CO2 ↑
-basically its happening in two steps, the carbonate diprotic acid forms then reacts again to produce water and co2


OR

CaCO3 + 2 HCl → CaCl2 + H2O + CO2 ↑

-this would be the overall reaction which greatly simplifies an otherwise somewhat confusing equation if you are new to stoichiometry

-if I just do the work for you it won't help you at all, I have a feeling your confusion is coming from determining the overall net reaction in balanced form so now that you have it, and i confirmed it with a reference, you should easily be able to cruise through the problem.

-just think of the basics, I can't remember the actual question but lets say you needto know how much CaCl2 will be made by reacting 10grams of CaCO3 with the amount of HCL needed to fully react with it.

10grams CaCO3*(#mols/10grams)=#of mols CaCO3 in ten grams CaCO3 =(X)

-you get 1 mol CaCl2 for every 1mol CaCo3 reacted so you get X mols of CaCl2

-Xmols of CaCl2 * ([molar mass CaCl2 in grams/mol] )= Y grams of CaCl2

-in Stoichiometry problems you must think limiting reagent which is the reactant that is present in the least amount on a relative molar scale. So if you have 5 mols CaCO3 and 9 mols HCl the limiting reactant is the HCl because 2 mols of HCl are required per mol of the carbonate so to fully react you would need 10 moles of HCl, therefore you would need to do the problem based on 9 mols of HCl, there would be a surplus of the carbonate as of course the products don't react with the carbonate or else they wouldn;t be products they would be reactants.



-there is an expression in boxing and martial arts "let your hands go", it can apply in a way to intellectual combat like learning chemistry. "let your mind go", try not to think to much and complicate that which may not be that complicated, just let you pencil start writing and work through the problem, get an answer, something to look at and analyse, just get something on paper and you'll find its much easier to break the problem apart and see where you are going wrong if you have something to look at.
-I know you are looking for more in the way of techincal help and I"m giving you a pep talk but really I think that is what you need. I think you feel a little beaten by being away from class and that can make you feel drowned, allow yourself to know the material and don't tell yourself you can't catch up. Take pride in attempting to learn one of the top five most difficult topics universities have to offer and that you are making a go of it, not many people can do that.
If you really need more of a walk through on your problems you can PM or email me at absolutely any time, I"ll give you my email on PM, don't ever feel like you are being a pain because i like the excercise. My background is bachelor in biochemistry and MSc in organic synthesis so early courses in university chemistry can be difficult for me when it comes to some theory but I'll do whatever i can to help a chemistry student, I've got all my old texts so its all in there.

[beaker]

Originally Posted by resmeplz

Balance the equation, figure out how much CaCO3 you have, then use mole ratios to figure out how much CaCO3 is converted to CO2.

For B you do it backwards, you know how much chalk and CO2 you have so figure out how much CaCO3 there is, then divide that by 4.38x100 to get the mass percent.

I tried to explain it the best I could ^^ if you still need help i can run and get some paper, just to lazy to walk outside to get it

after going back to look at the actual problem again i can say that this fellow heres response will get you the correct answer but remember that for professors and ta's they want to see neatly organized work

Grams CO2*moles CO2/gram CO2=
moles CO2*1mol CaCO3/1molCO2=molCaCO3
molCaCO3*grams CaCO3/molCaCO3=gramsCaCO3
(gramsCaCO3/4.38grams)*100%=percent of CaCO3 in the chalk sample

I hope I've answered the right question lol, I should have copied it into this response but the point i'm trying to make anyway is to show how you arrived at the answer by cancelling out units to arrive at the final unit, in this case %

just ignore the cosby douche, he only wrote that because it reminded him he doesn't know how to do it and probably never did, its just a way dick heads deflect attention from what ignorami they are. don't feel bad when someone who ate paste all through grades 1-7 just so no one would notice he had been urinating in his underoos

[Indigo]

Originally Posted by atara

We give you a hard time to encourage you to do as much as you can, not to scare you off.

>:3

Yea but dude said that after I had posted that I had solved it, there was simply no call for it....still I get it, and I certainly appreciate it.

So seldom do I ask for help (or post anything for that matter) that I often forget what a wonderful and helpful community you all have built here. If I ever come off as a wee bit hostile or defensive it is simply because my experience from other forums has made me that way.

With that shit outta the way....

Beaker, your advice of "let your hands go" is spot on my friend. I have learned that I have a bad habit of over thinking these kind of things, becoming overwhelmed and intimidated by the problem, but once I start writing it all kind of falls into place....

Thank you all for your advice and kind (and not so kind) words, with a bit of work, I hope to be able to one day contribute meaningfully to the community...you'll see, I'll do you all proud .

T.

[purewhitepanda]

E=mc2 is the answer, but if you would ask me I would tell you that E=getting fucked up mang......