posted by isabel .
if 2.00g of Zn is allowed to react with 2.00g of CuSO4 according to the equation CuSO4(aq) + Zn(s)-----> ZnSO4 (aq)+ Cu(s)
how many grams of Zn will remain after the reaction is complete ?
This is a limiting reagent problem. I know that because amounts are given for BOTH reactants. However since the problem asks for the amount of Zn remaining, one can assume that CuSO4 is the limiting reagent. I will show how we can confirm that.
CuSO4(aq) + Zn ==> ZnSO4(aq) + Cu(s)
mols CuSO4 = g/molar mass = 2.00/about 159.6 = 0.125
mols Zn = 2.00/65.4 = 0.0306
Convert mols of each to mols Cu produced.
0.0125 mols CuSO4 x (1 mol Cu/1 mol CuSO4) = 0.0125 mols Cu
0.0306 mols Zn x (1 mol Cu/1 mol Zn) = 0.0306 mols Cu
Of course both answers can't be correct; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that number is the limiting reagent. Therefore, 0.0125 mols Cu will be formed and CuSO4 is the limiting reagent. All of this just confirms what we knew at the beginning.
How much Zn remains unreacted? How many mols Zn were used? That will be
0.0125 mols Cu x (1 mol Zn/1 mol Cu) = 0.0125 mols Zn used.
mols Zn remaining = mols Zn initially-mols Zn used = 0.0306-0.0125 = ?
You can convert to grams by g Zn = mols Zn x atomic mass Zn.
mols Zn x (1 mul
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