Answer:
3.13 g of Fe remains after the reaction is complete
Explanation:
The first step to begin is determine the reaction:
2Fe + 3Cl₂ → 2FeCl₃
Let's find out the moles of each reactant:
22.7 g / 55.85 g/mol = 0.406 moles of Fe
37.2 g / 70.9 g/mol = 0.525 moles of Cl₂
Ratio is 2:3. 2 moles of iron react with 3 moles of chlorine
Then, 0.406 moles of iron will react with (0.406 . 3)/ 2 = 0.609 moles
We need 0.609 moles of chlorine when we have 0.525 moles, so as we do not have enough Cl₂, this is the limiting reactant.
The excess is the Fe. Let's see:
3 moles of chlorine react with 2 moles of Fe
Then, 0.525 moles of Cl₂ will react with (0.525 . 2) /3 = 0.350 moles
We need 0.350 moles of Fe and we have 0.406; as there are moles of Fe which remains after the reaction is complete, it is ok that Fe is the excess reagent.
0.406 - 0.350 = 0.056 moles of Fe still remains. We convert moles to mass:
0.056 mol . 55.85g / 1 mol = 3.13 g
