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

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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
 

Discussion

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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