Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: benworld on February 15, 2010, 02:06:34 PM
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Q: A mixture is 45.0 % NaCl (inert) and 55.0% Bacl2*2H20 . If 4.165 g of this mixture is heated until all of the hydrate is decomposed, what mass of solid residue will be left ?
My version:
Mass of Bacl2 = 208.20 g
Mass of 2H20 = 20 g
4.165g Bacl2*2H20
------------------ X 1 mole of Bacl2 = .20825 mol of Bacl2
20g of 2H2O
.20825 X 208.20 g of BaCl2
-------------------------- = 43.37 g of Bacl2
1 mole of bacl2
Does 43.27 g of Bacl2 left sound right ? I just can't figure it out.
Thanks
Ben
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correction to my calculation for molar mass of 2H20
how about know, does answer make sense..sorry for got to times oxygen Smiley
My version:
Mass of Bacl2 = 208.20 g
Mass of 2H20 = 36 g
4.165g Bacl2*2H20
------------------ X 1 mole of Bacl2 = .1156 mol of Bacl2
36g of 2H2O
.1156 X 208.20 g of BaCl2
-------------------------- = 24.08 g of Bacl2
1 mole of bacl2
Answer = 24.08 g of Bacl2 (s) solid residue left. This make sense ?
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Keep going, you're not there yet. (However, starting with 4.165 g and finishing with 24.08 g is impressive !!)
Seriously, estimation should tell you that the residual weight will be close to 4 g.
Motoball
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I'm confused, trying to do this problem for last 3 hours..1 problem...
How can I get good at problem like this ?
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okay, first you have to figure out what the decomposition reaction is.
Now I see a few problems.
Okay lets start at the beginning.
4.165g of NaCl + BaCl2+ 2H2O
55% of 4.165g is BaCl2+2H2O
So that means its 2.29075g of BaCl2+2H2O
Now you take
How did you get 36g for 2H2O??? You have to take 2 oxygen atoms which is 15.9994 each plus one hydrogen which is 1.00797 -- that means H2O weighs 33.00677 * 2 = 66.01354 g for 2H2O
Also how did you get 207 for BaCl2?? Ba is 137.77 + 70.906x2 = 279.142g
(2.29075g BaCl+2H2O)(1 mol BaCl2+2H2O/345.155554 BaCl+2H2O)(1 mol Ba/1 mol BaCl+2H2O)(137.77g Ba/1 mol Ba)
=0.91436056538149 grams of Ba
You should be able to figure out the rest.
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My bad on the H2O, lol I dono why I was thinking 2 oxygen, yours was right. BaCl2 appears to be different though.
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Not understanding how you did it.
1. The problem statement, all variables and given/known data
A mixture is 45.0 % of NaCl (inert) and 55.0 % BaCl2 * 2H20 . If 4.165 g of this mixture is heated until all of the hydrate is decomposed, what mass of solid residue will be left ?
2. Relevant equations
Na = 22.990
Cl = 35.453
Nacl = 58.443
Ba = 137.33
Cl = 2 x 35.45 = 70.906
H = 4 * 1.00 = 4.00
O = 2 x16.00 = 32.00 g
Bacl2*2h20 = 244.236
3. The attempt at a solution
1.
4.165 g Nacl(s)Bacl2*2H20 | 1 mole of Nacl + Bacl2*2H20
--------------------------------------------------------- = 0.0137 mol
| 302.679 g Nacl + Bacl2 * 2H20
2.
0.0137 mol Nacl+Bacl2*2H20 | 1 Mole of Nacl + Bacl2
------------------------------------------------ = 0.00685
| 2 Mole of 2H20
3.
0.00658 mol Nacl + Bacl2 | 266.679 Nacl + Bacl2 (mass)
----------------------------------------------------- = 1.75
| 1 mole of Nacl + Bacl2
Final Answer = 1.75 g of Nacl Bacl2
does this sound right ?
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Not really.
You seem to have assumed that it's 1:1 in terms of moles, among other things.
What mass of your original is NaCl? What mass is BaCl2.2H2O?
How many moles of BaCl2.2H2O do you have?
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Mass of Nacl = 58.443
Mass of Bacl2 * 2H20 = 244.236
Total Mass of Nacl + Bacl2 * 2H20 = 302.679
Nacl and Bacl2 have no mole.
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Same issue again.
If I had 100kg of apples, and 45% were green (the others being red), what mass of red apples would I have?
Now, if red apples weigh 45 g each, and green apples 110 g each, how many red and green apples do I have?
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I understand the analogie,but i can't seem to apply to current problem.
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I feel stupid, I'm sure this is so basic stuff
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Yes it is very simple basic stuff.
Lets go through it a step at a time.
You have 4.165g of a mixture.
That mixture is 45% NaCl and 55% BaCl2.2H2O
If you heat the mixture up the NaCl stays unchanged as NaCl but the BaCl2.2H2O losses the water to leave BaCl2
Of the 4.165g you stated with how much is BaCl2.2H2O in g
If that weight of BaCl2.2H2O losses the water to give BaCl2 how much does the BaCl2 weigh?
If you add the weight of BaCl2 to the weight of NaCl you started with that gives the correct answer for the final weight of dried solid.
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how about this
2.20975 g | 1 mol
------------------ = 0.00904 mol BaCl2*2H20
| 244.20 g
0.00904 mol | 2 mol H20
----------------------- = 0.01808
1 mol of Bacl2 * 2H20
0.01808 mol | 36 g
---------------- = 0.65088 g
1 mol
4.165 - 0.65088 g = 3.51412 g
?
Answer = 3.58644 g
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Not really.
You seem to have assumed that it's 1:1 in terms of moles, among other things.
What mass of your original is NaCl? What mass is BaCl2.2H2O?
How many moles of BaCl2.2H2O do you have?
Oh yeah I see what your saying, see I was assuming that it meant 55% BaCl2*2H2O by weight rather than moles. So in that case first you have to get the moles, then the percentage, then go back to grams.
So then benworld, first you have to get moles of NaCl & BaCl2*2H2O, then take the moles, put them into percentage, then calculate the weight.
For example (Im making up numbers here for example sake) if NaCL only weighed 0.0001 g per mol and BaCl2*2H2O weighed 1.0 g per mole, and 55% of the substance was BaCl2*2H2O then you would have very different weights between NaCl and BaCl*2H2O
5 mol of BaCl would be 5g and 5 mol of NaCl would be 0.0005 g. Once again I made up the numbers just to show you.
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thats is what i did
g to mol and mol to g
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Yea so just calculate it and there you go
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Nacl is inert elements, it has no effect at all.
1. Found Mol of Bacl2 * 2H20
2. Take the mol and times it by mole of H20 Since it H20 is being decomposed.
3. Take Mole of H20 and convert new substance into grams
4. Take the answer and minus it by original mass
It seems pretty simple as that ?
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Yea if you take the H20 weight and minus it from the original mixtures weight that will give you the weight of NaCl and BaCl2.
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Sounds..good..lol finally
36.00 g 2H20
------------ x 100 = 14% is how much dam 2H20 is
244.20
4.165 * 0.55 = 2.29075 g
2.29075 * 0.14 = 0.320705 g (2H20)
4.165 - 0.320705 = 3.844 g solid residu left ( bam...tell me if this is the answer ) :)
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Ah you cant do this 4.165 * 0.55 = 2.29075 g
Its 55% by mole not by weight right? So you have to figure out what the moles are, then calculate the percentage of weight of BaCl2*2H2O, then convert it back to grams, then calculate weight of 2H2O.
You know there should be formulas for this in your textbook..
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no its by weight...It says BaCl2 * 2H20 is 55 % ... So its assume its weight... aka....saying if entire sample is 4.156 g , then 55% of mixture 4.165 is 2.29075 in grams ?
Am I making sense.
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Oh lol ffs then it's rediculously easy. The other guy said it was by moles. Then yea you got it right as.
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Oh lol ffs then it's rediculously easy. The other guy said it was by moles. Then yea you got it right as.
Yeh but you still managed to make it more complicated by steering him in the wrong directions with your stupid answers for molecular weights. Also nobody but you said it was 45:55 in mole ratio, sjb simply pointed out benworld was wrong to use 1:1 mole or mass ratio. In future either make sure you are right or shut up.
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He ended up with the right answer didnt he? I misunderstood something, I apologize for being a fallible, imperfect human being.
Theres no need to be rude.
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Believe me this is not me being rude if I wanted to be rude you'd know about it.
Yes to ere is human but in this thread you have posted 8 times and of those 4 were incorrect and confused things. If you are going to post pointing out somebody else is wrong then the least you can do is be sure they are wrong and you are right. In your case the things you said were wrong were actually right and it was you who got it wrong. It would have been simple for you to look up the molecular weights of water and barium chloride to check who was right but you didn't bother. Those of us with more experience just know water is 18 etc. You being wrong in half the things you say is not a good track record for helping somebody else.