Fe(s) + CuSO4(aq) ==> FeSO4(aq) + Cu(s)
mols CuSO4 initially = M x L = 0.75 M x 0.100 L = 0.075
mols Fe initially = g/atomic mass = 2.5/55.85 = 0.045
This is a limiting reagent problem in which mols Fe are depleted completed and there is some CuSO4 remaining after the reaction ceases.
So how much Cu is formed. That is
0.045 mols Fe x (1 mol Cu/1 mol Fe) = 0.045 mols Cu produced.
grams Cu = 0.045 mols Cu x 63.54 g Cu/mol = 2.8 g Cu.
(FeSO4) = moles/L = 0.045 mols/0.100 L = ?
(CuSO4) = moles/L = ? You had 0.075 mols CuSO4 initally and you used 0.045 in the reaction so you have left 0.075-0.045 = 0.030
(CuSO4) = 0.030/0.100 = 0.30 M
2.5g of Iron Powder are added to 100ml of a 0.75mol/L Copper (II) Sulfate Solution.
a) determine the composition of the solution, in molar concentration when the reaction ceases?
b) What is the mass of copper produced?
1 answer
1 answer