Respuesta :
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
Answer:
Cl2 is the limiting reactant. Fe is in excess. There will remain 3.13 grams Fe
Explanation:
Step 1: data given
Mass of Fe = 22.7 grams
Atomic mass of Fe = 55.845 g/mol
Mass of Cl2 = 37.2 grams
Molar mass Cl2 = 70.9 g/mol
Step 2: The balanced equation
2Fe + 3Cl2 → 2FeCl3
Step 3: Calculate moles
Moles = Mass / molar mass
Moles Fe = 22.7 grams / 55.845 g/mol
Moles Fe = 0.406 moles
Moles Cl2 = 37.2 grams/ 70.9 g/mol
Moles Cl2 = 0.525 moles
Step 4: Calculate the limiting reactant
For 2 moles Fe we need 3 moles Cl2 to produce 2 moles FeCl3
Cl2 is the limiting reactant. It will completely be consumed (0.525 moles). Fe is in excess. There will react 2/3 * 0.525 = 0.35 moles
There will remain 0.406 - 0.350 = 0.056 moles
Step 5: Calculate mass of Fe remaining
Mass Fe = 0.056 moles* 55.845 g/mol
Mass Fe = 3.13 grams
Cl2 is the limiting reactant. Fe is in excess. There will remain 3.13 grams Fe
