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Chemistry Help! 10 points for best answer?
What is the mass of copper which is formed when excess aluminum is reacted with a given mass of copper(II) chloride dihydrate.
I've found the percent yield but I think my answer may be impossible. Instead of 100% I reached 142.5%.
8cm by 8cm aluminum foil, 2.00g copper(2) chloride dihydrate, 50 mL water.
3CuCl2 . 2H2O + 2Al -> 3 Cu + 2 AlCl3 + 6H2O, if I have 2 g of 3 CuCl2 . 2H2O, and the formula is
n= m/mm
n=2g/190.638 g/mol
n= 0.01049 mol
To find the mass of 3CuCl2 . 2H2O, do I add the 3 and the 2 in front of the compound and times 0.01049 mol by 3 or 5 or 6? I used 3 and my mass was 0.031473263 mol
m=n x mm
m= 0.031473263 mol x 63.546 g/mol
=2 g
My actual yield of copper was 2.85g, 163.15g of copper left over - 160.3g was the beakers weight = 2.85g of Cu
py= actual yield/expected yield x 100
=2.85g/2g x 100
=1.425g x 100
= 142.5%
Is that possible or did I make a mistake? Im not positive the materials I used were 100% pure, can that account for the large difference?
Please help, and thank you.
3 Answers
- 1 decade agoFavorite Answer
Of course it is possible you made a mistake! It happens to everyone.
I think your problem lies within your starting molecular mass. Your reaction is unbalanced.
CuCl2 . 2H2O is your molecule. When you balance them, and you realize it takes 3, then it takes 3 of all of it. So, calculate the molecular mass of CuCl2 . 2H2O and make your calculations based on that, not just the CuCl2 part.
- Anonymous4 years ago
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- Anonymous1 decade ago
Yep, it's perfectly possible. In some cases, you do end up with more product that you originally started with.
Help my friend?
