What mass of copper is formed when excess aluminum is reacted with a given mass of copper (II) chloride dihydrate?
3CuCl2 * 2H2O + 2Al --> 3Cu + 2AlCl3 + 6H2O
1.00 g
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letsbefree
Canada123 Posts
What mass of copper is formed when excess aluminum is reacted with a given mass of copper (II) chloride dihydrate? 3CuCl2 * 2H2O + 2Al --> 3Cu + 2AlCl3 + 6H2O 1.00 g | ||
Lemonwalrus
United States5465 Posts
(AM of Cu) + 2(AM of Cl) + 4(AM of H) + 2(AM of O) = Molecular Weight of Copper (II) chloride dihydrate. Since we know that there is 1 gram of the Copper (II) chloride dihydrate: 1 Gram / MW calculated above = moles of Copper (II) chloride dihydrate. Since moles of Copper (II) chloride dihydrate in the reactants = moles of Copper in the products just multiply the number you get above by the atomic mass of copper, and that should be your answer. Haven't had chem in a while though, might want to double check. | ||
Lemonwalrus
United States5465 Posts
Edit: Nvmd, edited. | ||
fight_or_flight
United States3988 Posts
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Lemonwalrus
United States5465 Posts
On March 27 2008 10:59 fight_or_flight wrote: Just find the total mass of copper in the left side. Do this by finding the mass of 3 moles of copper, 6 moles of chloride, etc. Then you have the ratio of each element. Then just multiply the ratio of copper mass by 1 gram. Did my way work and this is just faster, or is my way wrong? | ||
fight_or_flight
United States3988 Posts
Also, I said the left side, but I don't think you should include aluminum. | ||
letsbefree
Canada123 Posts
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fight_or_flight
United States3988 Posts
On March 27 2008 11:08 letsbefree wrote: i'm a bit confused guys, do i treat the water portion of the hydrate as one entire mol or 2 mols? 3CuCl2 * 2H2O you have here 3 Cu masses, 6 Cl masses, 4 H masses, and 2 O masses. Adding all those individual masses together gives you the total mass of the molecule. Since you have 1 gram of this molecule, you can find the number of molecules, which gives you the number of copper atoms. | ||
Lemonwalrus
United States5465 Posts
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Lemonwalrus
United States5465 Posts
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GrayArea
United States872 Posts
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Lemonwalrus
United States5465 Posts
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letsbefree
Canada123 Posts
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Nitrogen
United States5345 Posts
3 CuCl2 * 2H2O + 2Al --> 3Cu + 2AlCl3 + 6H2O 1.00 g (Copied so I can read while replying) First, you must find the molar mass of copper (II) chloride dihydrate. (170.389 g/mol) Next, set up an equation. (1.00 g CuCl2 * 2 H20 / 170.389 g/mol CuCl2 * 2 H20) x (63.54 g/mol Cu) Because the coefficients of CuCl2 * 2 H20 and Cu are the same in the equation, you can disregard them. The answer comes out to be 0.373 g Cu. | ||
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