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17 July, 19:47

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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  1. 17 July, 20:46
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    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
  2. 17 July, 22:02
    0
    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
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