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Old 03-19-2012, 06:22 PM   #41
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your gonna need a periodic table for stoichiometry. you are also going to need to find the limiting reactant. hopefully somebody can make it clear how to do that because i am not great at it. once you find that you do this.

take the initial mass of whatever you are starting with. for example, you have 54.0g of Al. you then need to find the molar mass or how much 1 mole of Al weighs(the number usually underneath of Al on the periodic table) then you take the number of moles you start with which in this case is 1 which you know from the equation. then you take the number of moles of Cu that will be made in this case 3 and you also need the molar mass of Cu. now once you know all that set it up like this

initial mass(g)*(number of mol of reactant/molar mass reactant(g))*( number of mol of product/number of mol of reactant)*(molar mass of product(g))= total mass of product(g)

its hard to show on here but if you set it up so when you divide it is literally 1 over molar mass you will be able to see that units will cancel out and that is how you know you did it right. I don't want to give you the answer because it isnt really helping anyone if i do haha but let me know if you want me to clear anything up.
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Old 03-19-2012, 06:29 PM   #43
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Originally Posted by 113tac View Post
your gonna need a periodic table for stoichiometry. you are also going to need to find the limiting reactant. hopefully somebody can make it clear how to do that because i am not great at it. once you find that you do this.

take the initial mass of whatever you are starting with. for example, you have 54.0g of Al. you then need to find the molar mass or how much 1 mole of Al weighs(the number usually underneath of Al on the periodic table) then you take the number of moles you start with which in this case is 1 which you know from the equation. then you take the number of moles of Cu that will be made in this case 3 and you also need the molar mass of Cu. now once you know all that set it up like this

initial mass(g)*(number of mol of reactant/molar mass reactant(g))*( number of mol of product/number of mol of reactant)*(molar mass of product(g))= total mass of product(g)

its hard to show on here but if you set it up so when you divide it is literally 1 over molar mass you will be able to see that units will cancel out and that is how you know you did it right. I don't want to give you the answer because it isnt really helping anyone if i do haha but let me know if you want me to clear anything up.
Thanks dude! I think what I'm confused on is how to set up this part "initial mass(g)*(number of mol of reactant/molar mass reactant(g))*( number of mol of product/number of mol of reactant)*(molar mass of product(g))= total mass of product(g)" just where and why you place each part where you do.
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Old 03-19-2012, 06:51 PM   #45
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i made a picture haha. basically 1 mol= molar mass so those are basically a conversion factor. then the mol of product/mol of reactant are to convert between the compounds. it is like a proportion i guess. here is a picture



you multiply across both top and bottom and then divide the top by the bottom. showed how they cancel out too color coated and all! haha
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Old 03-19-2012, 06:57 PM   #46
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basically, you put

the weight/quantity given in the equation, in this case

54.0g Al

then multiply/divide that by the molar mass of the reactant, in this case

54.0g Al 1mol Al / 27g

then multiply/divide that by the number of mols of the reactant, compared to the number of mols of the product. Both can be found in the original equation, in this case

54.0gAl 1mol Al/27g 3mol Cu/ 1mol Al

work that through and you end up with 6 mols of Cu made



do the same for CuSO4, and find which number is smaller.


the one with the smaller number is the limiting reactant; that reactant will get used up before the other, and how much product is made depends on that limiting reactant.

then once you have how many mols of the product is made with the limiting reactant, just multiply the mols made by the molar mass.
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Old 03-19-2012, 07:20 PM   #47
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Originally Posted by Konaborne View Post
basically, you put

the weight/quantity given in the equation, in this case

54.0g Al

then multiply/divide that by the molar mass of the reactant, in this case

54.0g Al 1mol Al / 27g

then multiply/divide that by the number of mols of the reactant, compared to the number of mols of the product. Both can be found in the original equation, in this case

54.0gAl 1mol Al/27g 3mol Cu/ 1mol Al

work that through and you end up with 6 mols of Cu made



do the same for CuSO4, and find which number is smaller.


the one with the smaller number is the limiting reactant; that reactant will get used up before the other, and how much product is made depends on that limiting reactant.

then once you have how many mols of the product is made with the limiting reactant, just multiply the mols made by the molar mass.
How did you get 6mols of Cu? I keep getting 0.666??
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Old 03-20-2012, 07:09 AM   #48
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Ok let's break this down a little.

First, you need to make sure you balance that equation!!

Since your answer isn't affected for this first question, you'd get away with it, but I'd prefer you know how to use your tools.

The equation (balanced) is

2Al + 3CuSO4 ~~> Al2(SO4)3 + 3 Cu

First, get your molecular/atomic weights together. My advice is, for the purpose of practice, to do them all, even if you don't need them. It helps alot both for your understanding and for your teacher to see that you understand it to show your work. Write it all out--they'll be able to follow your logic and you'll have an easier time pinpointing when you've made a mistake.

Al = 26.981g/mol

CuSO4 =

Cu=63.546
S=32.065
O=15.999 (x4)

CuSO4 = 63.546 + 32.065 + (15.999x4) = 159.607g/mol

So now we can figure out how many moles we have to work with:

1 mol Al x 54g Al = 2.001 mol Al
26.981g Al

1 mol CuSO4 x 319g CuSO4 = 1.9987 mol CuSO4
159.607g CuSO4

Written out this way you can be sure visually that the units cancel.

We've got 1.9987 mol CuSO4. We want to know how many moles of Cu we can make with that.

3 mol Cu x 1.9987 mol CuSO4 = 1.9987 mol Cu
3 mol CuSO4

The 3's cancel, as do the "mol CuSO4" leaving you with 1.9987 mol Cu.

So now we simply convert the moles of Cu to grams using the atomic weight of Cu, which we've already written out (see again why it's good to write it all out?)

63.546g Cu x 1.9987 mol Cu = 127.009g Cu
mol Cu
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Old 03-20-2012, 08:02 AM   #49
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It's been awhile since I've done the gas stoichiometry thing, but after poking around a little, I've jogged my memory enough to remember that with Standard Temperature and Pressure, this is a very easy thing to figure out.

First, make sure you have a balanced equation:

2H2 + O2 ~~> 2H2O(g)

(I've specified here that the product is supposed to be in gas form. If you were collecting the water as a liquid, the beginning would be the same, but you would end a little differently. I can do both just for giggles here.)

First, convert to moles (moles moles moles!! It's all about the moles. If you can convert various reactants and products to moles, you're 90% of the way to understanding what you're doing!)

For gases, you simply use the Ideal Gas Law (the complicated version is a calculation using PV=nRT ~~Pressure x Volume = moles of gas x gas constant x temperature (K)

The shortened version (made SO easy with the statement "STP) is that at Standard Temperature and Pressure, 1 mole of gas = 22.4 Liters of gas. If you hated your life enough (I sure as hell don't, I own a Tacoma ) you could use the above equation and work it all out and get the same answer.

So long story short here, back to converting to moles:

H2:
1 mol H2 x 8.6L H2 =0.384 mol H2
22.4L H2

O2:
1 mol O2 x 4.3L O2 = 0.192 mol O2
22.4L O2

Interestingly, these come out exactly (they are in exact proportions, so you can use either reactant to find your product), since you need 2 moles of H2 for 1 mole O2 and 0.192 (mol O2) x 2 is 0.384 (mol H2)

To get moles H2O:

Testing above statement, I'll do it using each reactant.

H2:

2 mol H2O x 0.384mol H2 = 0.384 mol H2O
2 mol H2

O2:

2 mol H2O x 0.192mol O2 = 0.384 mol H2O
1 mol O2

To convert moles H2O in gas form back to liters, we use the same conversion factor we used for the other gases:

22.4L H2O x 0.384mol H2O = 8.6L H2O (g)
1 mol H2O

***

If you were to collect this as a liquid, you would use the molecular weight of the water instead (to produce the product in grams):

18.0g H2O x 0.384mol H2O = 6.912g H2O (l)
1 mol H2O
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Old 03-20-2012, 08:25 AM   #50
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im a chemistry major in my senior year. so i will check back on this ASAP. im actually being a wonderful studious individual right now and browsing TW while in my government class. but will get right back on here to see if i can help.
GodSpeed
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Old 03-20-2012, 08:27 AM   #51
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This gives me nightmares. Studying in high school was a non-issue for me...except for chemistry! I think that was my only C and I was glad to have it. Best of luck.
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Old 03-20-2012, 08:55 AM   #52
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Originally Posted by UNDEAD1153 View Post
I believe I'm gonna have to take you up on this. Ive been working problems for about 2 hours and I still have no idea...this chapter is stoichiometry. A couple questions:

Metallic copper is formed when aluminum reacts with copper(II) sulfate. How many grams of metallic copper can be obtained when 54.0g of Al react with 319g of CuSO4?
Al+3CuSO4--->Al2(SO4)3 + 3Cu


If 8.6L of H2 reacted with 4.3L of O2 at STP, what is the volume of the gaseous water collected?
2H2 + O2----> 2H2O


How many liters of O2 are needed to react completely with 45.0L of H2S at STP?
2H2S + 3O2-----> 2SO2 + 2H2O


If anyone can explain to me what to do and why you are doing it I would greatly appreciate it!

Study!
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Old 03-20-2012, 08:55 AM   #53
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Originally Posted by m3bassman View Post
Will someone do my diff eq HW for me?
Yeah, I will as soon as I am done taking that class.......
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Old 03-20-2012, 10:32 AM   #54
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Originally Posted by SCGuyInACOMA90 View Post
im a chemistry major in my senior year. so i will check back on this ASAP. im actually being a wonderful studious individual right now and browsing TW while in my government class. but will get right back on here to see if i can help.
GodSpeed

Wanna help me with Ochem??
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Old 03-20-2012, 11:06 AM   #55
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Quote:
Originally Posted by Squisha View Post
It's been awhile since I've done the gas stoichiometry thing, but after poking around a little, I've jogged my memory enough to remember that with Standard Temperature and Pressure, this is a very easy thing to figure out.

First, make sure you have a balanced equation:

2H2 + O2 ~~> 2H2O(g)

(I've specified here that the product is supposed to be in gas form. If you were collecting the water as a liquid, the beginning would be the same, but you would end a little differently. I can do both just for giggles here.)

First, convert to moles (moles moles moles!! It's all about the moles. If you can convert various reactants and products to moles, you're 90% of the way to understanding what you're doing!)

For gases, you simply use the Ideal Gas Law (the complicated version is a calculation using PV=nRT ~~Pressure x Volume = moles of gas x gas constant x temperature (K)

The shortened version (made SO easy with the statement "STP) is that at Standard Temperature and Pressure, 1 mole of gas = 22.4 Liters of gas. If you hated your life enough (I sure as hell don't, I own a Tacoma ) you could use the above equation and work it all out and get the same answer.

So long story short here, back to converting to moles:

H2:
1 mol H2 x 8.6L H2 =0.384 mol H2
22.4L H2

O2:
1 mol O2 x 4.3L O2 = 0.192 mol O2
22.4L O2

Interestingly, these come out exactly (they are in exact proportions, so you can use either reactant to find your product), since you need 2 moles of H2 for 1 mole O2 and 0.192 (mol O2) x 2 is 0.384 (mol H2)

To get moles H2O:

Testing above statement, I'll do it using each reactant.

H2:

2 mol H2O x 0.384mol H2 = 0.384 mol H2O
2 mol H2

O2:

2 mol H2O x 0.192mol O2 = 0.384 mol H2O
1 mol O2

To convert moles H2O in gas form back to liters, we use the same conversion factor we used for the other gases:

22.4L H2O x 0.384mol H2O = 8.6L H2O (g)
1 mol H2O

***

If you were to collect this as a liquid, you would use the molecular weight of the water instead (to produce the product in grams):

18.0g H2O x 0.384mol H2O = 6.912g H2O (l)
1 mol H2O
Thanks so much for helping! I took the test (sorta....) in my third period, only 2 people out of 19 finished, we all have to go back after school and finish. I'm still beyond confused and trying to learn by 3:00 EST.... Might be impossible....
I'm not understanding why you put what you get where you do because there are so many different types of problems and no set equation to use. If there is an equation I can use for the different types I could memorize those but otherwise I don't know what to do, it's just not clicking....
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Old 03-20-2012, 11:29 AM   #56
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If you can convert your reactants/products to moles, you're 75% of the way there. Do you understand how to do that?

If you can demonstrate that step 1 is to make sure the equation is balanced, and step 2 is to convert everything to moles, you're probably going to scrape enough partial credit (assuming your teacher isn't a hard-case) to pass.

Is your teacher an old guy? I'm sorry, I guess I'm biased, from personal experience. Old guys don't seem able to relate to their students (regardless of age, really) well enough to be able to break things down and build them back up. Then they have unreasonable expectations at exam time.

If 90% of the class was unable to finish in the time allotted, then clearly his expectations were unreasonable.
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Old 03-20-2012, 11:54 AM   #57
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this is blowing my mind every way possible.

good job on helping though, i tried to read and learn but idk what im even reading haha
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Old 03-20-2012, 12:51 PM   #58
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Originally Posted by Squisha View Post
If you can convert your reactants/products to moles, you're 75% of the way there. Do you understand how to do that?

If you can demonstrate that step 1 is to make sure the equation is balanced, and step 2 is to convert everything to moles, you're probably going to scrape enough partial credit (assuming your teacher isn't a hard-case) to pass.

Is your teacher an old guy? I'm sorry, I guess I'm biased, from personal experience. Old guys don't seem able to relate to their students (regardless of age, really) well enough to be able to break things down and build them back up. Then they have unreasonable expectations at exam time.

If 90% of the class was unable to finish in the time allotted, then clearly his expectations were unreasonable.
Not exactly....is it just dividing the given by the amu? My teacher is probably in his late 40's.....his help for us is "if you don't know it, know it" so I have no idea what to do, I also missed a full week of the chapter due to pneumonia. But good news is I passed last 9 weeks with a C thanks to y'all! Bad thing is that I'm probably looking at a 20 on the test because I have no idea how to do the equations and that's all the test is....
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Old 03-20-2012, 05:58 PM   #59
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Bump to save my grade!
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Old 03-20-2012, 06:26 PM   #60
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when does this test happen? good luck kid. i wrestled with it, but i got it finally..

i never used it again in the real world.

kudos for asking for divine intervention on a chem exam..
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