Question

When iron is reacted with aqueous iron(III) ions, iron(II) ions are formed. Assuming the reaction goes to completion, how many moles of Fe and of Fe3+ (aq) would result in a mixture containing equal numbers of moles of Fe3+(aq) and Fe2+(aq) once the reaction had taken place? moles of Fe moles of Fe3+ (aq) 2 A B 3 C 1 5 D 2 3​

Answer 1

Explanation:

for the reaction between metallic iron and a solution of copper (II) sulfate. This reaction

produces metallic copper, which is seen precipitating as a finely divided red powder. This type

of reaction, in which one metal "displaces" another from a solution of one of its salts, is known

as a single substitution reaction. A metal capable of displacing another from a solution of one of

its salts is said to be "more active" than the displaced metal. In this experiment, iron is more

active than copper.

Iron forms 2 types of ions, namely Fe+2 and Fe+3. We shall use stoichiometric principles to

determine which of these ions is formed in the reaction between iron and copper (II) sulfate

solution. If Fe+2 is formed, then equation (1) is correct, while equation (2) is correct if Fe+3 is

formed. Your task is to find out which equation is consistent with the results of your experiment.

(1) Fe(s) + CuSO4 (aq) ! FeSO4 (aq) + Cu(s)

Fe(s) + Cu+2(aq) ! Fe+2(aq) + Cu(s)

(2) 2Fe(s) + 3CuSO4 (aq) ! Fe2(SO4)3 (aq) + 3Cu(s)

2Fe(s) + 3Cu+2(aq) ! 2Fe+3(aq) + 3Cu(s)

An excess of copper (II) sulfate solution (to make sure that all the iron is reacted) will be

added to a known amount of iron. The metallic copper produced will be weighed. These

weighings will be used to calculate the moles of iron used and the moles of copper formed. If

equation (1) is correct, the moles of copper should equal the moles of iron. If equation (2) is

correct, we should obtain 1.5 moles of copper per mole of iron.