Final answer:
To prepare 11.86 g of iron chloride, approximately 4.08 g of iron is needed.
Step-by-step explanation:
In order to determine the mass of iron needed to prepare 11.86 g of iron chloride, we need to use stoichiometry. The balanced equation for the reaction is:
2 Fe + 3 Cl2 → 2 FeCl3
From this equation, we can see that the molar ratio between iron and iron chloride is 2:2. This means that one mole of iron reacts with one mole of iron chloride. To find the mass of iron needed, we can use the molar mass of iron chloride:
Molar mass of FeCl3: 162.2 g/mol
Now we can set up a proportion to solve for the mass of iron:
(11.86 g FeCl3) / (162.2 g/mol) = (x g Fe) / (55.8 g/mol)
Solving for x, we find that the mass of iron needed is approximately 4.08 g.