Final answer:
To find how much FeCl3 should form when 35.0g of iron reacts with 35.0g of chlorine, stoichiometry is used to determine that chlorine is the limiting reactant, resulting in 53.53g of FeCl3 being produced.
Step-by-step explanation:
To determine how much FeCl3 should form when 35.0g of iron reacts with 35.0g of chlorine, we need to perform a stoichiometric calculation based on the balanced chemical equation.
First, calculate the moles of iron and chlorine using their molar masses (Fe = 55.85 g/mol, Cl2 = 70.90 g/mol):
- Moles of Fe = 35.0 g / 55.85 g/mol = 0.63 mol
- Moles of Cl2 = 35.0 g / 70.90 g/mol = 0.49 mol
Then, use the balanced chemical equation to find the limiting reactant, which will determine the maximum amount of FeCl3 that can be formed: 2 Fe(s) + 3 Cl2(g) ——> 2 FeCl3(s).
From the stoichiometry of the reaction, we can see that 2 mol of Fe reacts with 3 mol of Cl2. However, we have 0.63 mol of Fe and only 0.49 mol of Cl2, meaning Cl2 is the limiting reactant.
Now, calculate the maximum amount of FeCl3 based on the amount of Cl2:
- 0.49 mol Cl2 x (2 mol FeCl3 / 3 mol Cl2) = 0.33 mol FeCl3
Finally, convert the moles of FeCl3 to grams using its molar mass (FeCl3 = 162.20 g/mol): 0.33 mol FeCl3 x 162.20 g/mol = 53.53 g of FeCl3.
Therefore, if 35.0g of iron reacts with 35.0g of chlorine, 53.53g of FeCl3 should form, assuming complete reaction and no losses.