Check my Answer - Let me know if its right or wrong

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Discussion Overview

The discussion revolves around a homework problem involving the calculation of the mass of solid residue remaining after heating a mixture of NaCl and BaCl2·2H2O. Participants explore the use of mass percentages and stoichiometry to arrive at the final mass after decomposition.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Ben presents an initial calculation attempting to find the mass of solid residue left after heating the mixture, but does not incorporate the given mass percentages of the components.
  • Some participants emphasize the importance of using the mass percentages to determine the individual masses of NaCl and BaCl2·2H2O in the mixture.
  • One participant suggests calculating the moles of BaCl2·2H2O based on its mass and then determining the mass of water released during decomposition.
  • Another participant questions the calculation of the mass of water, pointing out a potential error in the molecular weight used for water.
  • Subsequent calculations are presented by participants, showing different approaches to arrive at the mass of solid residue, with some corrections noted in the process.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach or final answer, as multiple methods and calculations are proposed, leading to different results.

Contextual Notes

Participants express uncertainty regarding the application of mass percentages and the calculations of moles and mass, highlighting potential errors in earlier steps without resolving them.

benworld
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Homework Statement



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 ?


Homework Equations



1 mole substance = grams of substance
1 mole of substance = 6.022E23 of substance ( avogardo number)
Nacl = 58.443
Bacl2*2H20 = 244.236

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 ?


Thanks
Ben
 
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Did you use any of those % you are given, they are important.
 
No, How do I use them in formula ? is my setup wrong ?
 
% are usually by mass if not specified otherwise.

So of the 4.165g 45% is NaCl and 55%is BaCl2. Considering that you can find out the mass of salt and the mass of hidrated BaCl2. From there you get the moles then the weight of the water contained and finally by substracting the mass of water from the initial weight you get your answer.
 
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
 
benworld said:
how about this

2.20975 g 4.165*0.55=2.29075 | 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 Why 36 h20 has 18!
---------------- = 0.65088 g
1 mol

4.165 - 0.65088 g = 3.51412 g

?

Answer = 3.58644 g
Some corrections above
 

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