Identify the limiting reactant in the reaction of iron and chlorine to form FeCl3, of 22.7 g of Fe and 37.2 g of Cl2 are combined. Determine the amount (in grams) of excess reactant that remains after the reaction is complete.

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

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

the cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.