Posted by 496935sd on Friday, March 26, 2010 at 6:58pm.
I think the preferred way of writing CuCO3 is copper(II) carbonate and not copper(2) carbonate.
Cu(s) ==>CuCO3==>CuSO4.5H2O.
One mole of Cu is involved from beginning to end; therefore, moles Cu = 5.463 g/atomic mass Cu = ??
That will give you ?? moles (the same number) of CuSO4.5H2O.
Grams CuSO4.5H2O = moles CuSO4.5H2O x molar mass CuSO4.5H2O = xx. This is the theoretical yield.
%yield = (actual yield/theoretical yield)*100 = zz.
You will need to follow the directions. I don't know why you are to do the stoichiometry twice. You didn't write the hint so I don't know if there are instructions there, too. Finally, remember to round to the correct number of significant figures.
ooo srry the hint was that u have to do the method twice
and thank u soo much
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