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### Topic: Question...Check my Answer  (Read 14065 times)

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#### benworld

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• Mole Snacks: +0/-0 « on: February 15, 2010, 02:06:34 PM »
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

#### benworld

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• Mole Snacks: +0/-0 « Reply #1 on: February 15, 2010, 02:45:15 PM »
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 ?

#### MOTOBALL

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• Mole Snacks: +43/-5 « Reply #2 on: February 15, 2010, 04:52:40 PM »
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

#### benworld

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• Mole Snacks: +0/-0 « Reply #3 on: February 15, 2010, 05:09:09 PM »
I'm confused, trying to do this problem for last 3 hours..1 problem...

How can I get good at problem like this ?

#### maximus242

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• Mole Snacks: +0/-1 « Reply #4 on: February 15, 2010, 09:02:56 PM »
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.

#### maximus242

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• Mole Snacks: +0/-1 « Reply #5 on: February 15, 2010, 10:09:27 PM »
My bad on the H2O, lol I dono why I was thinking 2 oxygen, yours was right. BaCl2 appears to be different though.

#### benworld

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• Mole Snacks: +0/-0 « Reply #6 on: February 16, 2010, 08:19:59 AM »
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 ?

#### sjb

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• Gender:  « Reply #7 on: February 16, 2010, 08:29:25 AM »
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?

#### benworld

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• Mole Snacks: +0/-0 « Reply #8 on: February 16, 2010, 08:46:36 AM »
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.

#### sjb

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• Gender:  « Reply #9 on: February 16, 2010, 09:15:33 AM »
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?

#### benworld

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• Mole Snacks: +0/-0 « Reply #10 on: February 16, 2010, 09:43:29 AM »
I understand the analogie,but i can't seem to apply to current problem.

#### benworld

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• Mole Snacks: +0/-0 « Reply #11 on: February 16, 2010, 09:43:54 AM »
I feel stupid, I'm sure this is so basic stuff

#### DrCMS

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• Gender:  « Reply #12 on: February 16, 2010, 12:29:03 PM »
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.

#### benworld

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• Mole Snacks: +0/-0 « Reply #13 on: February 16, 2010, 01:21:35 PM »

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

?

#### maximus242

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• Mole Snacks: +0/-1 « Reply #14 on: February 16, 2010, 01:29:59 PM »
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.