Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: JonathanEyoon on February 16, 2008, 05:53:46 PM
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Water can be formed from the stoichiometric reaction of hydrogen with oxygen:
A complete reaction of 5.0 g of O2 with excess hydrogen produces how many grams of H2O?
I'm having a worst time with this one than the other one I had. I don't even know where to start off ??? Any help is much appreciated.
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Dear JonathanEyoon;
“Every” Stoichiometry Problem starts with the same Steps:
- A.) Build the “Reaction Equation” for your Problem.
- B.) Balance1) the “Reaction Equation” with the required Indices, giving you the Stoichiometry.
- C.) Translate the known masses for the “Reaction Equation” into the corresponding number of Moles.
- D.) Maybe you have to find the limiting Reagent.
- E.) Calculate from C.) and D.) the still missing masses using the number of Moles from the “Reaction Equation”.
- F.) Identify your Result.
1) For “How to Balance a Chemical Reaction” you may read on: "Balancing Chemical Equations (http://richardbowles.tripod.com/chemistry/balance.htm)”
2) How the ‘whole’ Diagram/Scheme for the second Example must look you can see on the: "Replay #36 and #37 (http://www.chemicalforums.com/index.php?topic=23025.msg88010#msg88010)”.
I hope this small Recipe may be of help to you to solve your Stoichiometry.
Good Luck!
ARGOS++
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I don't even know how to set up the problem ??? That's why i'm having such a hard time with this stuff.
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Start with rxn eqn (as Argos++ mentioned)
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ok. How do I start that?
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Dear JonathanEyoon;
Let’s try to do A.) and B.) together:
A.) Reaction: O2 + H2 -----> H2O.
B.) Balancing: 1O2 + 2H2 -----> 2H2O.
C.) Are you able to find all required Molecular Weights (MW), and translate your 5.0 g O2 into the Number of Moles?
Will this be enough for the Start?
Good Luck!
ARGOS++
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yea I know how to find MW and convert to mols. What do I do from there?
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Use mole ratios - every mole of O2 gives 2 mole of H2O.
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Dear JonathanEyoon;
As C1.) Write below the Reaction Members the corresponding MWs, and
As C2.) Write the translated Moles of 5.0g O2 below its MW.
For D.) you know that the Amount of O2 is the only limiting Reagent.
From E.) and B.) you know how many moles water you can build from one Mole O2, and so you can calculate the Amount in Moles and Grams of Water you have build.
Good Luck!
ARGOS++
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??? i'm completely lost. Is there any way you could do this one problem for me? I want to see an example of how to do a problem of this kind. I have 14 more after this one :-\ Any help is much appreciated.
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yea I know how to find MW and convert to mols. What do I do from there?
Great. You're given amounts in units of grams. The balanced chemical equation is in units of moles. You want the answer in grams.
So convert grams of reactant to moles, use the balanced equation to convert moles of reactant into moles of product (hey, why do you think we call it a balanced equation? It is a function that you've derived, just like mathematics, for the purpose of converting moles of product into moles of reactants.) Then convert moles of product back into grams, for the answer.
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??? i'm completely lost. Is there any way you could do this one problem for me? I want to see an example of how to do a problem of this kind. I have 14 more after this one :-\ Any help is much appreciated.
At least show us that you can determine molecular weight, and convert grams to moles, like you assured us you can. Then try to follow the steps outlined by ARGOS++. Sev's suggestion is good as well, but you may need a bit more expertise, before his suggestion becomes intuitive, and not confusing for you.
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Dear JonathanEyoon;
Don’t fear!
From the indices given on B.) you know that from 1 Mole O2 you can build 2 Moles Water.
So Please tell me how many Moles of O2 you have (= 5.0g), and how many Moles of Water you have built.
Good Luck!
ARGOS++
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Wait, how do we get 2 H20's from 1 mole of 02? Don't we need a hydrogen for that to happen? 5.0g of oxygen to mols should be
.3125 moles
How do you find how many moles of water i've built?
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Dear JonathanEyoon;
Sorry!, ─
I got 0.15625 Moles, because the MW of O2 is 32.0 and not 16.0!
With an Excess of H2 you can build 2 * 0.15625 Moles Water (from B.)).
How many Water will this now be?
Good Luck!
ARGOS++
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Dear JonathanEyoon;
Sorry!, ─
I got 0.15625 Moles, because the MW of O2 is 32.0 and not 16.0!
With an Excess of H2 you can build 2 * 0.15625 Moles Water (from B.)).
How many Water will this now be?
Good Luck!
ARGOS++
So 2 x 0.15625 is gonna be .3125 moles of water?
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Dear JonathanEyoon;
Exactly!, ─ And how many Grams will that finally be?
Good Luck!
ARGOS++
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also in simple terms, what does it mean that their is an excess of H2? Excess usually means it's left over. Which should mean it should have no bearing at all to the problem right?
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Dear JonathanEyoon;
Exactly!, ─ And how many Grams will that finally be?
Good Luck!
ARGOS++
So the final answer to this problem is 5.625 grams?
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also in simple terms, what does it mean that their is an excess of H2? Excess usually means it's left over. Which should mean it should have no bearing at all to the problem right?
Yes, either they give you a stoichiometric amount, or something is in excess.
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Dear JonathanEyoon;
Bravo, You did it!
Yes, an Excess means that there is more then enough from that stuff, and so its amount will in no kind influence the Amount of the Product.
(Please remember the Recipe!)
Good Luck!
ARGOS++
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I still don't understand how to do these types of problems ??? Can we try another one?
How about this one.
If 294 grams of FeS2 is allowed to react with 176 grams of O2 according to the following equation, how many grams of Fe2O3 are produced?
The equation: FeS2 + 02 -----> Fe203 + SO2
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Dear JonathanEyoon;
Will you try to do A.) and B.) for this Reaction?
Good Luck!
ARGOS++
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Dear JonathanEyoon;
Will you try to do A.) and B.) for this Reaction?
Good Luck!
ARGOS++
Ok so part A and B is to write the equation and balance it out right?
So the balanced equation should be
4FeS2 + 11O2 ------> 2Fe2O3 + 8SO2? What should I do from here?
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Great, you know the units of your balanced equation are given in moles, you've been given reactants in units of grams. You have to convert. Have you been given both reactants in stoichiometric amounts, or is one in excess?
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Dear JonathanEyoon;
(Hello again.)
Very well done!, ─
You know after A.) and B.), now C.) must follow:
So we need for C1.) all MW’s, and you write it just below each Educt and Product of your well found “Reaction Equation”.
After that you write for C2. ) all known masses of the Educts, but now translated into the corresponding Number of Moles under the MW’s.
Would you like to tell me the results of your line C2. )?
Good Luck!
ARGOS++
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Dear JonathanEyoon;
(Hello again.)
Very well done!, ─
You know after A.) and B.), now C.) must follow:
So we need for C1.) all MW’s, and you write it just below each Educt and Product of your well found “Reaction Equation”.
After that you write for C2. ) all known masses of the Educts, but now translated into the corresponding Number of Moles under the MW’s.
Would you like to tell me the results of your line C2. )?
Good Luck!
ARGOS++
Ok so 294 grams would be for FeS2 and 176 grams would be for the O2. So i change these two into moles? If that is so the
294 grams FeS2 = 294 / 120 = 2.45 Moles of FeS2
176 Grams O2 = 176 / 32 = 5.5 moles of O2
where do I go from here ???
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Dear JonathanEyoon;
Excelent!, ─
Now you have to find the limiting Reagent and that we do in line D.).
For that you must “put” your calculated Moles from line C2. ) on the same Denominator (to the same Basis) using the found indices from B.).
For Example: Divide your 2.45 Moles by 4 (the Indices from B.) for FeS2), and so on.
The lowest Number on line D.) tells you the limiting Reagent; with other words: All other Reagents of the Equation are in Excess and so you can calculate from the Moles of the limiting Reagent how many Moles of your asked Product you can build. Use again the Indices from line B.).
Now you have only to translate the Number of Moles back to Grams.
Do you think you can make the Desission and try to calculate ahead?
(I think I need now an eye full of sleep.)
Good Luck!
ARGOS++
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Dear JonathanEyoon;
Excelent!, ─
Now you have to find the limiting Reagent and that we do in line D.).
For that you must “put” your calculated Moles from line C2. ) on the same Denominator (to the same Basis) using the found indices from B.).
For Example: Divide your 2.45 Moles by 4 (the Indices from B.) for FeS2), and so on.
The lowest Number on line D.) tells you the limiting Reagent; with other words: All other Reagents of the Equation are in Excess and so you can calculate from the Moles of the limiting Reagent how many Moles of your asked Product you can build. Use again the Indices from line B.).
Now you have only to translate the Number of Moles back to Grams.
Do you think you can make the Desission and try to calculate ahead?
(I think I need now an eye full of sleep.)
Good Luck!
ARGOS++
haha I'm not too sure how to proceed with this ???
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Ok, i think I got the answer. Is the answer 98 Grams?
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No, too low.
You have 2.45mol FeS2 and 5.5mol O2. You have to find which reagent is limiting - consumed first.
From the rxn eqn 4 mole FeS2 reacts with 11 mole O2. That means 2 moles of FeS2 reacts with 5.5 moles of O2 (not 2.45!). So O2 is the limiting reagent (it is completely consumed in the reaction).
Now, look at the coefficients from the rxn eqn. How many moles of Fe2O3 are formed from 5.5 mole of O2?
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so what should the answer be? I'm asking because this way i can work backwards to towards something and try to understand it myself while doing it. I'm still so lost on how to do this stuff with the explanations on here. Can't understand some of the stuff you guys are saying. So I'm gonna try to work towards the answer or backwards from it and try to figure out a simple way to do this stuff since i'm sure their is a way method to do all of this. Thanks and hope to hear it soon
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You end up with 1 mole of Fe2O3. Do you know what a limiting reagent is?
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no idea what that is
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The limiting reagent is completely consumed in the rxn. In this case, it is O2.
FeS2 is in excess - there is still some left after all of O2 has reacted.
See: http://www.chembuddy.com/?left=balancing-stoichiometry&right=limiting-reagents :)
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First of all - on the same site - read how to read balanced reaction equation and what it means:
http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations
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Dear JonathanEyoon;
Let’s collect together in the Diagram what you have calculated till jet:
A.) Reaction: FeS2 + O2 -----> Fe2O3 + SO2.
B.) Balancing1): 4FeS2 + 11O2 -----> 2Fe2O3 + 8SO2.
C1.) MW’s: 120.0 g/m 32.0 g/m 158.7 g/m 64.0 g/m
Cm.) Masses: 294.0 g 176.0 g
C2.) Moles: 2.45 m 5.50 m
Now I was asking for the Decision following exactly my Example above:
You should result in: 2.45 m / 4 = 0.6125 ; and -
5.50 m / 11 = 0.5000.
As you don’t know any other Masses or Moles of any Reactant, your line D.) must look like:
D.) Multiplier: 0.6125 0.5000
Now it should be clear that you can only produce the Amount, what the smallest Multiplier telling you, all other is in Excess. All Excess can NOT react, because it has NO “Reaction Partner”!
So you can copy the smallest Multiplier to all your missing/(not known till yet) Multipliers of your “Unknowns” on line D. ). That must result in:
D.) Multiplier: 0.6125 0.5000 0.50 0.50
The final step is now to complete lines C2.) and Cm.) for all “Unknowns” from the bottom up, because you have figured out the Multipliers for all.
That gives you:
C2.) Moles: 2.45 m 5.5 m 2 * 0.50 m 8 * 0.50m.
And to get the final masses you have only to back-translate the new values from line C2.) into masses, to complete also line Cm.) as you did it for the water in the first Example.
Yet you know how much of each Product you were able to produce.
1) For “How to Balance a Chemical Reaction” you may read on: "Balancing Chemical Equations (http://richardbowles.tripod.com/chemistry/balance.htm)”
Will you try to complete your Diagram this way and tell me your final results?
(Finally we put the whole Diagram together.)
Good Luck!
ARGOS++
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Dear JonathanEyoon;
Maybe you’re anyway coming once back, but it’s also for all who try to learn from this Topic!
(I also like to “hold” my contracts.)
Here is the resulting Diagram for your Stoichiometry Problem:
A.) Reaction: FeS2 + O2 -----> Fe2O3 + SO2.
B.) Balancing: 4FeS2 + 11O2 -----> 2Fe2O3 + 8SO2.
C1.) MW’s: 120.0 g/m 32.0 g/m 158.7 g/m 64.0 g/m
Cm.) Masses: 294.0 g 176.0 g 158.7 g 256.0 g
C2.) Moles: 2.45 m 5.50 m 2 * 0.50 m 8 * 0.50 m
D.) Multiplier: 0.6125 0.5000 0.50 0.50
As you can see: All Mass results of you Stoichiometry Problem are located on the Cm.) line.
Surprise: Also all Excesses can be calculated in the same easy manner:
Add on the left an additional Column for each Excess and calculate from the bottom to the top.
For your particular Stoichiometry Problem it looks like:
A.) Reaction: Excess (FeS2)
B.) Balancing:
C1.) MW’s: 120.0 g/m
Cm.) Masses: 54.0 g
C2.) Moles: 4 * 0.1125 m
D.) Multiplier: 0.1125 (= 0.6125 – 0.5000)
Now the Diagram gives you the “whole” Overview about your Stoichiometry Problem and can easy be controlled/validated, because the line Cm.) must end in identity for the left and the right side, except all (multiplied) small rounding errors.
(Remember: Excesses you have to subtract on the left Side!)
I hope it may be of help.
Good Luck!
ARGOS++
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http://www.chembuddy.com/?left=balancing-stoichiometry&right=ratio-proportions
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Dear Mr.Borek;
Bravo! “Chembuddy.com”, ─ Quite a good Page with good Explanations.
But it doesn’t answers the initial Stoichiometry Problems, and it covers only a very small part of the Stoichiometry Problems at all.
It doesn’t handle “Limiting Reagent” of Stoichiometry Problems.
It doesn’t address/calculate “Excesses” of Stoichiometry Problems.
Also the control/validation of the calculations of the “whole” Stoichiometry Problems is missing.
So I still believe that the Diagram is the smallest, simplest, and most complete calculation of such and most other related Stoichiometry Problems.
And it gives you at once the “whole” Overview about your Stoichiometry Problem.
But the Page is a good start point for beginners of Stoichiometry Problems.
Good Luck!
ARGOS++
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But it doesn’t answers the initial Stoichiometry Problems, and it covers only a very small part of the Stoichiometry Problems at all.
http://www.chembuddy.com/?left=balancing-stoichiometry&right=toc
It doesn’t handle “Limiting Reagent” of Stoichiometry Problems.
http://www.chembuddy.com/?left=balancing-stoichiometry&right=limiting-reagents
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Dear Mr. Borek;
I beg your Pardon!, ─ I did not recognise the last Link on the left of your first presented Page. (Maybe I took the wrong glasses, Sorry!)
But the Diagram has still some advantages, - but your Pages have much more Explanations!
Sorry once again.
Good Luck!
ARGOS++