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.

Question

Maxine

Chemistry

5 August, 19:21

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.

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Answers (2)

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Brennen Solis · Answer 1

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

Trent Harrell · Answer 2

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